[proofplan]
We prove the assertion locally in holomorphic coordinates, because membership in $\Omega^{p,q}(M)$ is defined by the local expression of a form in the basis generated by $dz_I \wedge d\bar z_J$. A form of type $(p,q)$ is a finite sum of smooth coefficients times $dz_I \wedge d\bar z_J$ with $|I|=p$ and $|J|=q$. Complex conjugation conjugates the coefficients and sends each $dz_i$ to $d\bar z_i$ and each $d\bar z_j$ to $dz_j$. Reordering the wedge factors introduces only the sign $(-1)^{pq}$, so every conjugated term has type $(q,p)$.
[/proofplan]
[step:Write the form in a holomorphic coordinate basis]
Let $(U,z)$ be a holomorphic coordinate chart on $M$, where
\begin{align*}
z: U \to z(U) \subseteq \mathbb{C}^n
\end{align*}
has coordinate functions $z_1,\dots,z_n$. For an increasing multi-index $I=(i_1,\dots,i_p)$, define
\begin{align*}
dz_I := dz_{i_1}\wedge \cdots \wedge dz_{i_p}.
\end{align*}
For an increasing multi-index $J=(j_1,\dots,j_q)$, define
\begin{align*}
d\bar z_J := d\bar z_{j_1}\wedge \cdots \wedge d\bar z_{j_q}.
\end{align*}
Since $\alpha \in \Omega^{p,q}(M)$, the local-coordinate characterization of forms of type $(p,q)$ gives smooth coefficient functions
\begin{align*}
a_{I,J}: U \to \mathbb{C}
\end{align*}
such that, on $U$,
\begin{align*}
\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} a_{I,J}\, dz_I\wedge d\bar z_J.
\end{align*}
[/step]
[step:Conjugate each local basis term and reorder it into standard type]
By definition, complex conjugation on complex-valued differential forms is complex-antilinear, conjugates the coefficient functions, and satisfies
\begin{align*}
\overline{dz_i}=d\bar z_i
\end{align*}
and
\begin{align*}
\overline{d\bar z_i}=dz_i
\end{align*}
for each $i\in\{1,\dots,n\}$. Therefore, on $U$,
\begin{align*}
\bar\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} \overline{a_{I,J}}\, d\bar z_I\wedge dz_J.
\end{align*}
Each term $d\bar z_I\wedge dz_J$ is reordered by moving the $q$ factors in $dz_J$ past the $p$ factors in $d\bar z_I$. Each interchange of one $1$-form with another contributes a factor of $-1$, so the total sign is $(-1)^{pq}$. Hence
\begin{align*}
d\bar z_I\wedge dz_J = (-1)^{pq} dz_J\wedge d\bar z_I.
\end{align*}
Thus
\begin{align*}
\bar\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} (-1)^{pq}\overline{a_{I,J}}\, dz_J\wedge d\bar z_I.
\end{align*}
[guided]
The point of this step is to check exactly what conjugation does to the two kinds of coordinate $1$-forms. The convention for complex-valued forms is that conjugation is complex-antilinear: coefficients are conjugated, and the real exterior algebra part is conjugated by sending the complex coordinate form $dz_i$ to $d\bar z_i$ and the conjugate coordinate form $d\bar z_i$ back to $dz_i$. Therefore the local expression
\begin{align*}
\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} a_{I,J}\, dz_I\wedge d\bar z_J
\end{align*}
becomes
\begin{align*}
\bar\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} \overline{a_{I,J}}\, d\bar z_I\wedge dz_J.
\end{align*}
This expression has the right factors but in the opposite order from the standard convention for a form of type $(q,p)$, which is $dz_J\wedge d\bar z_I$. To put it into standard order, move the $q$ factors of $dz_J$ past the $p$ factors of $d\bar z_I$. Since exterior multiplication of $1$-forms is alternating, each pairwise swap contributes the factor $-1$. There are $pq$ such swaps, so
\begin{align*}
d\bar z_I\wedge dz_J = (-1)^{pq} dz_J\wedge d\bar z_I.
\end{align*}
Substituting this into the expression for $\bar\alpha|_U$ gives
\begin{align*}
\bar\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} (-1)^{pq}\overline{a_{I,J}}\, dz_J\wedge d\bar z_I.
\end{align*}
The sign is harmless because it is a scalar factor and can be absorbed into the smooth coefficient.
[/guided]
[/step]
[step:Conclude that the conjugate has type $(q,p)$ on every chart]
For each pair $(I,J)$, the coefficient function
\begin{align*}
b_{J,I}: U \to \mathbb{C}
\end{align*}
defined by
\begin{align*}
b_{J,I}:=(-1)^{pq}\overline{a_{I,J}}
\end{align*}
is smooth, because complex conjugation of a smooth complex-valued function is smooth. The local expression
\begin{align*}
\bar\alpha|_U=\sum_{\substack{|I|=p, |J|=q}} b_{J,I}\, dz_J\wedge d\bar z_I
\end{align*}
is therefore a finite sum of smooth coefficients multiplying basis forms with $q$ holomorphic factors and $p$ antiholomorphic factors. Since the chart $(U,z)$ was arbitrary, $\bar\alpha$ is locally of type $(q,p)$ everywhere on $M$. Hence
\begin{align*}
\bar\alpha\in\Omega^{q,p}(M).
\end{align*}
This proves the theorem.
[/step]
