[proofplan] We associate to each $(P,\omega)$-partition $\sigma$ a unique linear extension by sorting the elements of $P$ first by the value of $\sigma$ and then by their label. This produces a weakly increasing sequence of values, and descents in the resulting label word force strict increases. Conversely, a linear-extension label word together with such a sequence defines a map $P\to\mathbb N$, and the linear-extension property plus the descent strictness condition verify exactly the two defining inequalities for a $(P,\omega)$-partition. Finally, the tie-breaking construction recovers the original label word, so the parametrization is disjoint and exhaustive. [/proofplan] [step:Sort a $P$-partition by value and then by label] Let $\sigma:P\to\mathbb N$ be a $(P,\omega)$-partition. Since $P$ is finite, there is a unique ordering $x_1,\dots,x_n$ of the elements of $P$ such that the pairs \begin{align*} (\sigma(x_i),\omega(x_i)) \end{align*} are weakly increasing in lexicographic order. Define the associated label word $\pi_\sigma=(\pi_1,\dots,\pi_n)$ by \begin{align*} \pi_i=\omega(x_i) \end{align*} for every $i\in\{1,\dots,n\}$. We prove that $\pi_\sigma\in\mathcal L(P,\omega)$. Let $x,y\in P$ satisfy $x<y$. Because $\sigma$ is a $(P,\omega)$-partition, we have $\sigma(x)\le\sigma(y)$. If $\sigma(x)<\sigma(y)$, then the lexicographic sort places $x$ before $y$. If $\sigma(x)=\sigma(y)$, the strict part of the $(P,\omega)$-partition condition rules out $\omega(x)>\omega(y)$; hence $\omega(x)<\omega(y)$ because $\omega$ is injective, and the tie-break by increasing label again places $x$ before $y$. Thus every comparable pair is ordered correctly, so $\pi_\sigma$ is the label word of a linear extension. [guided] The sorting rule is designed to encode both inequalities in the definition of a $(P,\omega)$-partition. We define an ordering $x_1,\dots,x_n$ of $P$ by requiring the pairs $(\sigma(x_i),\omega(x_i))$ to be weakly increasing in lexicographic order. This means that smaller $\sigma$-values come first, and among equal $\sigma$-values the smaller label comes first. Because $\omega:P\to\{1,\dots,n\}$ is bijective, no two distinct elements have the same label, so this tie-breaking gives a unique order. Now define $\pi_\sigma=(\pi_1,\dots,\pi_n)$ by $\pi_i=\omega(x_i)$. To show that $\pi_\sigma$ is a linear-extension label word, we must prove that whenever $x<y$ in $P$, the element $x$ appears earlier than $y$ in the sorted order. Since $\sigma$ is a $(P,\omega)$-partition, the weak defining inequality gives \begin{align*} \sigma(x)\le\sigma(y). \end{align*} If $\sigma(x)<\sigma(y)$, then $x$ appears before $y$ because the sort is increasing in the first coordinate. If $\sigma(x)=\sigma(y)$, then the strict defining condition says that $\omega(x)>\omega(y)$ would force $\sigma(x)<\sigma(y)$, which contradicts $\sigma(x)=\sigma(y)$. Therefore $\omega(x)>\omega(y)$ is impossible. Since $\omega$ is injective and $x\ne y$, we must have $\omega(x)<\omega(y)$. The tie-break by increasing label then places $x$ before $y$. Thus every relation $x<y$ in $P$ is respected, so the sorted order is a linear extension and $\pi_\sigma\in\mathcal L(P,\omega)$. [/guided] [/step] [step:Read off the sequence and prove the descent strictness condition] Define a sequence $(a_1,\dots,a_n)\in\mathbb N^n$ by \begin{align*} a_i=\sigma(x_i)=\sigma(\omega^{-1}(\pi_i)) \end{align*} for every $i\in\{1,\dots,n\}$. Since the elements were sorted by nondecreasing $\sigma$-value, we have \begin{align*} 1\le a_1\le \cdots \le a_n. \end{align*} Let $i\in\operatorname{Des}(\pi_\sigma)$. Then $\pi_i>\pi_{i+1}$, so $\omega(x_i)>\omega(x_{i+1})$. If $a_i=a_{i+1}$, the lexicographic sorting rule would put the smaller label first, contradicting $\omega(x_i)>\omega(x_{i+1})$. Hence $a_i<a_{i+1}$. Therefore the sequence attached to $\sigma$ satisfies the required descent strictness condition. [/step] [step:Build a $P$-partition from a linear extension and an admissible sequence] Conversely, let $\pi=(\pi_1,\dots,\pi_n)\in\mathcal L(P,\omega)$. Let $(a_1,\dots,a_n)\in\mathbb N^n$ satisfy \begin{align*} 1\le a_1\le \cdots \le a_n \end{align*} and \begin{align*} i\in\operatorname{Des}(\pi)\implies a_i<a_{i+1}. \end{align*} Define the map $\sigma_{\pi,a}:P\to\mathbb N$ by \begin{align*} \sigma_{\pi,a}(\omega^{-1}(\pi_i))=a_i \end{align*} for every $i\in\{1,\dots,n\}$. This is well-defined because $\omega$ is bijective and $\pi$ is a permutation of $\{1,\dots,n\}$. We verify the $(P,\omega)$-partition inequalities. Let $x,y\in P$ satisfy $x<y$. Since $\pi$ is a linear-extension label word, there exist indices $i<j$ such that $x=\omega^{-1}(\pi_i)$ and $y=\omega^{-1}(\pi_j)$. The weak monotonicity of the sequence gives \begin{align*} \sigma_{\pi,a}(x)=a_i\le a_j=\sigma_{\pi,a}(y). \end{align*} Now assume also that $\omega(x)>\omega(y)$. Then $\pi_i>\pi_j$. Since the sequence of labels moves from $\pi_i$ to the smaller value $\pi_j$ over the adjacent positions $i,i+1,\dots,j$, there is some $k\in\{i,\dots,j-1\}$ such that $\pi_k>\pi_{k+1}$. Thus $k\in\operatorname{Des}(\pi)$, so $a_k<a_{k+1}$. Combining this strict inequality with the weak inequalities along the sequence gives \begin{align*} a_i<a_j. \end{align*} Therefore \begin{align*} \sigma_{\pi,a}(x)<\sigma_{\pi,a}(y). \end{align*} So $\sigma_{\pi,a}$ is a $(P,\omega)$-partition. [/step] [step:Show the two constructions are inverse and the union is disjoint] Starting with a $(P,\omega)$-partition $\sigma$, the preceding construction produces $\pi_\sigma$ and $a_i=\sigma(\omega^{-1}((\pi_\sigma)_i))$. The assignment reconstructed from this pair satisfies \begin{align*} \sigma_{\pi_\sigma,a}(\omega^{-1}((\pi_\sigma)_i))=a_i=\sigma(\omega^{-1}((\pi_\sigma)_i)) \end{align*} for every $i\in\{1,\dots,n\}$, and these elements exhaust $P$. Hence the reconstructed assignment is exactly $\sigma$. Conversely, begin with $\pi\in\mathcal L(P,\omega)$ and an admissible sequence $(a_1,\dots,a_n)$. Let $x_i=\omega^{-1}(\pi_i)$ for every $i$. The sequence of pairs \begin{align*} (\sigma_{\pi,a}(x_i),\omega(x_i))=(a_i,\pi_i) \end{align*} is lexicographically weakly increasing: if $a_i<a_{i+1}$ this is immediate, and if $a_i=a_{i+1}$ then $i\notin\operatorname{Des}(\pi)$, so $\pi_i<\pi_{i+1}$. Therefore sorting $\sigma_{\pi,a}$ by value and then by label recovers the same order $x_1,\dots,x_n$ and hence the same label word $\pi$. Thus each $(P,\omega)$-partition determines exactly one linear-extension label word and one admissible sequence, and each such pair determines exactly one $(P,\omega)$-partition. The parametrizing pieces indexed by distinct $\pi\in\mathcal L(P,\omega)$ are therefore disjoint, and their union is the set of all $(P,\omega)$-partitions. This proves the theorem. [/step]