[proofplan]
We realize weakly order-preserving maps as lattice points in dilates of the order polytope of $P$. The shift from the target set $\{1,\dots,m\}$ to $\{0,\dots,m-1\}$ identifies $\Omega_P(m)$ with the Ehrhart counting function of the closed order polytope at dilation factor $m-1$. Ehrhart reciprocity then converts the negative evaluation into a count of interior lattice points in the dilation by $m+1$. Finally, those interior lattice points are exactly the strictly order-preserving maps $P\to\{1,\dots,m\}$.
[/proofplan]
[step:Identify weakly order-preserving maps with lattice points in a shifted order polytope]
If $P=\varnothing$, then $n=0$, there is exactly one map $P\to\{1,\dots,m\}$ for every positive integer $m$, and the same unique map is strictly order-preserving. Since the order polynomial agrees with the constant function $1$ on all positive integers, we have $\Omega_P(t)=1$ in $\mathbb{Q}[t]$, and hence $\Omega_{P,\mathrm{str}}(m)=1=(-1)^0\Omega_P(-m)$. Thus assume for the rest of the proof that $P\neq\varnothing$.
Let $\mathbb{R}^P$ denote the real [vector space](/page/Vector%20Space) of functions $u:P\to\mathbb{R}$. Define the order polytope of $P$ by
\begin{align*}
\mathcal{O}(P)=\{u\in\mathbb{R}^P:0\leq u(x)\leq 1\text{ for every }x\in P,\text{ and }u(x)\leq u(y)\text{ whenever }x<y\}.
\end{align*}
By [citetheorem:8086], the finite poset $P$ has a linear extension. Choose a linear extension and write it as a bijection $\ell:P\to\{1,\dots,n\}$ satisfying $x<y\implies \ell(x)<\ell(y)$. Define $u_0:P\to\mathbb{R}$ by $u_0(x)=\ell(x)/(n+1)$. Then $0<u_0(x)<1$ for every $x\in P$, and $u_0(x)<u_0(y)$ whenever $x<y$. Since $P$ is finite, the positive number
\begin{align*}
\delta=\min\Bigl(\{u_0(x):x\in P\}\cup\{1-u_0(x):x\in P\}\cup\{u_0(y)-u_0(x):x<y\}\Bigr)
\end{align*}
is well-defined. If $u:P\to\mathbb{R}$ satisfies $|u(x)-u_0(x)|<\delta/3$ for every $x\in P$, then $0<u(x)<1$ for every $x\in P$ and $u(x)<u(y)$ whenever $x<y$. Thus $\mathcal{O}(P)$ contains a nonempty open ball in the ambient vector space $\mathbb{R}^P$, so $\dim \mathcal{O}(P)=\dim \mathbb{R}^P=n$.
We record that $\mathcal{O}(P)$ is a lattice polytope. Let $v\in\mathcal{O}(P)$ be a vertex. Suppose that some coordinate $v(x)$ lies strictly between $0$ and $1$. Define an undirected graph $G_v$ on the vertex set $P$ by joining distinct comparable elements $a,b\in P$ exactly when $v(a)=v(b)$. Let $C\subset P$ be the connected component of $x$ in $G_v$. Along every edge of $G_v$, equality of the endpoint coordinates forces the same infinitesimal displacement on both endpoints, so any perturbation preserving all active comparable-pair equalities must be constant on $C$.
If no element $a\in C$ satisfies $v(a)=0$ or $v(a)=1$, then every active equality involving a coordinate in $C$ is either one of the comparable-pair equalities defining the edges of $G_v$ or an equality internal to the common perturbation of $C$. Because $P$ is finite, all inactive inequalities have a positive margin. Hence for sufficiently small $\varepsilon>0$, the two maps $v_+,v_-:P\to\mathbb{R}$ defined by $v_\pm(a)=v(a)\pm\varepsilon$ for $a\in C$ and $v_\pm(a)=v(a)$ for $a\notin C$ both remain in $\mathcal{O}(P)$. Since $v=(v_++v_-)/2$ and $v_+\neq v_-$, this contradicts that $v$ is a vertex. Therefore $C$ is pinned by an active equality $v(a)=0$ or $v(a)=1$ for some $a\in C$. Since $v$ is constant on $C$, the common value on $C$ is $0$ or $1$, contradicting $0<v(x)<1$. Hence every vertex has coordinates in $\{0,1\}$, so every vertex belongs to $\mathbb{Z}^P$.
Let $\mathbb{Z}_{\geq 0}$ denote the set of nonnegative integers. Define the Ehrhart counting map
\begin{align*}
L_{\mathcal{O}(P)}:\mathbb{Z}_{\geq 0}\to\mathbb{Z}_{\geq 0},\qquad k\mapsto |k\mathcal{O}(P)\cap\mathbb{Z}^P|,
\end{align*}
where $\mathbb{Z}^P$ is the lattice of integer-valued maps $P\to\mathbb{Z}$. A point $g\in k\mathcal{O}(P)\cap\mathbb{Z}^P$ is exactly an order-preserving map $g:P\to\{0,\dots,k\}$.
For a positive integer $m$, define a map between two finite sets by
\begin{align*}
\Phi_m:\{f:P\to\{1,\dots,m\}:f\text{ is order-preserving}\}\to (m-1)\mathcal{O}(P)\cap\mathbb{Z}^P,\qquad f\mapsto f-1.
\end{align*}
Its inverse sends $g$ to $g+1$. Hence
\begin{align*}
\Omega_P(m)=L_{\mathcal{O}(P)}(m-1).
\end{align*}
[/step]
[step:Apply Ehrhart reciprocity with the correct dilation shift]
By [citetheorem:8134], $\Omega_P(m)$ agrees on positive integers with a polynomial in $m$. Since $\mathcal{O}(P)$ is a lattice polytope, Ehrhart's polynomial theorem gives a unique polynomial, also denoted $L_{\mathcal{O}(P)}(t)\in\mathbb{Q}[t]$, whose value at every nonnegative integer $k$ is $|k\mathcal{O}(P)\cap\mathbb{Z}^P|$. The preceding step gives $\Omega_P(m)=L_{\mathcal{O}(P)}(m-1)$ for every positive integer $m$. Since two polynomials that agree on infinitely many integers are equal, the order polynomial is the shifted Ehrhart polynomial
\begin{align*}
\Omega_P(t)=L_{\mathcal{O}(P)}(t-1).
\end{align*}
Therefore, for a positive integer $m$,
\begin{align*}
\Omega_P(-m)=L_{\mathcal{O}(P)}(-m-1).
\end{align*}
We now use the Ehrhart reciprocity theorem for the $n$-dimensional lattice polytope $\mathcal{O}(P)$: if $Q\subset\mathbb{R}^n$ is a lattice polytope of dimension $n$ with Ehrhart polynomial $L_Q(t)$, then for every positive integer $r$ one has $L_Q(-r)=(-1)^n|rQ^\circ\cap\mathbb{Z}^n|$. The hypotheses match here because the preceding step proved that $\mathcal{O}(P)$ has dimension $n$ in $\mathbb{R}^P$ and that all vertices of $\mathcal{O}(P)$ lie in the lattice $\mathbb{Z}^P$. Therefore, for every positive integer $r$,
\begin{align*}
L_{\mathcal{O}(P)}(-r)=(-1)^n |r\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|,
\end{align*}
where $\mathcal{O}(P)^\circ$ denotes the interior of $\mathcal{O}(P)$ in $\mathbb{R}^P$. Applying this with $r=m+1$ gives
\begin{align*}
(-1)^n\Omega_P(-m)=|(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|.
\end{align*}
[guided]
We first redeclare the objects used in this reciprocity step. The ambient vector space is $\mathbb{R}^P$, the lattice is $\mathbb{Z}^P$, and the order polytope is
\begin{align*}
\mathcal{O}(P)=\{u\in\mathbb{R}^P:0\leq u(x)\leq 1\text{ for every }x\in P,\text{ and }u(x)\leq u(y)\text{ whenever }x<y\}.
\end{align*}
Let $\mathbb{Z}_{\geq 0}$ denote the set of nonnegative integers. Its Ehrhart counting function is the map
\begin{align*}
L_{\mathcal{O}(P)}:\mathbb{Z}_{\geq 0}\to\mathbb{Z}_{\geq 0},\qquad k\mapsto |k\mathcal{O}(P)\cap\mathbb{Z}^P|.
\end{align*}
A point of $k\mathcal{O}(P)\cap\mathbb{Z}^P$ is exactly an order-preserving map $P\to\{0,\dots,k\}$. Therefore the translation $f\mapsto f-1$ gives a bijection from order-preserving maps $P\to\{1,\dots,m\}$ to $k\mathcal{O}(P)\cap\mathbb{Z}^P$ with $k=m-1$. Thus, for every positive integer $m$,
\begin{align*}
\Omega_P(m)=L_{\mathcal{O}(P)}(m-1).
\end{align*}
Both sides agree with polynomials in the integer parameter: $\Omega_P$ does so by [citetheorem:8134], and Ehrhart's polynomial theorem gives a unique polynomial extension $L_{\mathcal{O}(P)}(t)\in\mathbb{Q}[t]$ once $\mathcal{O}(P)$ is known to be a lattice polytope. Hence equality on all positive integers gives the polynomial identity
\begin{align*}
\Omega_P(t)=L_{\mathcal{O}(P)}(t-1).
\end{align*}
Evaluating at $t=-m$ gives
\begin{align*}
\Omega_P(-m)=L_{\mathcal{O}(P)}(-m-1).
\end{align*}
We now verify the hypotheses needed for Ehrhart reciprocity. The polytope is full-dimensional in $\mathbb{R}^P$: because $P\neq\varnothing$, a linear extension $\ell:P\to\{1,\dots,n\}$ exists by [citetheorem:8086], and the map $u_0:P\to\mathbb{R}$ defined by $u_0(x)=\ell(x)/(n+1)$ satisfies $0<u_0(x)<1$ for all $x\in P$ and $u_0(x)<u_0(y)$ whenever $x<y$. Since $P$ is finite, the positive margins in these strict inequalities have a positive minimum, so a sufficiently small open ball around $u_0$ in $\mathbb{R}^P$ is contained in $\mathcal{O}(P)$. Thus $\dim \mathcal{O}(P)=\dim\mathbb{R}^P=n$.
The polytope is also a lattice polytope. Indeed, if $v\in\mathcal{O}(P)$ is a vertex and some coordinate $v(x)$ lies strictly between $0$ and $1$, form the graph on $P$ whose edges connect comparable pairs $a<b$ with $v(a)=v(b)$. On the connected component of $x$, every active comparable-pair equality forces a common perturbation. If that component is not pinned by an active equality $v(a)=0$ or $v(a)=1$, then all coordinates on the component may be moved by a sufficiently small common amount both upward and downward; finiteness of $P$ gives a positive margin for every inactive inequality, so both perturbed points remain in $\mathcal{O}(P)$. This writes $v$ as the midpoint of two distinct points of $\mathcal{O}(P)$, contradicting that $v$ is a vertex. If the component is pinned by $v(a)=0$ or $v(a)=1$, then the common value on the component is $0$ or $1$, again contradicting $0<v(x)<1$. Hence every coordinate of every vertex is $0$ or $1$, and every vertex belongs to $\mathbb{Z}^P$.
Ehrhart reciprocity applies to the full-dimensional lattice polytope $\mathcal{O}(P)\subset\mathbb{R}^P$. It gives, for every positive integer $r$,
\begin{align*}
L_{\mathcal{O}(P)}(-r)=(-1)^n |r\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|,
\end{align*}
where $\mathcal{O}(P)^\circ$ denotes the interior in $\mathbb{R}^P$. Taking $r=m+1$ and using the shifted identity for $\Omega_P(-m)$ yields
\begin{align*}
\Omega_P(-m)=(-1)^n |(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|.
\end{align*}
Multiplying by $(-1)^n$ gives
\begin{align*}
(-1)^n\Omega_P(-m)=|(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|.
\end{align*}
[/guided]
[/step]
[step:Identify interior lattice points with strictly order-preserving maps]
Let $h:P\to\mathbb{Z}$ be an integer-valued map. The condition
\begin{align*}
h\in (m+1)\mathcal{O}(P)^\circ
\end{align*}
is equivalent to
\begin{align*}
0<h(x)<m+1\text{ for every }x\in P,\qquad h(x)<h(y)\text{ whenever }x<y.
\end{align*}
Indeed, the nontrivial defining inequalities $0\leq u(x)\leq 1$ and $u(x)\leq u(y)$ for comparable pairs $x<y$ become strict in the interior. Conversely, if all these strict inequalities hold, then $h/(m+1)$ satisfies every nontrivial defining inequality with positive margin, so it lies in $\mathcal{O}(P)^\circ$.
Since $h$ is integer-valued, the inequalities $0<h(x)<m+1$ are equivalent to $h(x)\in\{1,\dots,m\}$ for every $x\in P$. Thus
\begin{align*}
(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P=\{h:P\to\{1,\dots,m\}:h(x)<h(y)\text{ whenever }x<y\}.
\end{align*}
The right-hand side is precisely the set counted by $\Omega_{P,\mathrm{str}}(m)$. Therefore
\begin{align*}
|(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|=\Omega_{P,\mathrm{str}}(m).
\end{align*}
[/step]
[step:Combine the Ehrhart count with the strict map count]
From the Ehrhart reciprocity step,
\begin{align*}
(-1)^n\Omega_P(-m)=|(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|.
\end{align*}
From the interior-point identification,
\begin{align*}
|(m+1)\mathcal{O}(P)^\circ\cap\mathbb{Z}^P|=\Omega_{P,\mathrm{str}}(m).
\end{align*}
Combining these equalities gives
\begin{align*}
\Omega_{P,\mathrm{str}}(m)=(-1)^n\Omega_P(-m).
\end{align*}
This is the desired reciprocity formula for every positive integer $m$.
[/step]
