Let $X$, $Y$ and $Y'$ be (standard) Borel spaces. We let $\mathcal B(X)$ be the Borel $\sigma$-algebra of $X$ and $\mathcal P(X)$ to be the space of all Borel probability distributions on $X$ endowed with the topology of weak convergence, so that it is a Borel space as well. Consider any Borel set $A\in \mathcal B(X\times \mathcal P(Y))$ and a pushforward operator $\pi_*:\mathcal P(Y\times Y')\to \mathcal P(Y)$ defined as $$ (\pi_*p)(B) = p(B\times Y') \qquad \forall B\in \mathcal B(Y). $$ What can we say about measurability of $A':=\{(x,p):\pi_*p\in A_x\}\subseteq X\times \mathcal P(Y\times Y')$, where $$ A_x:=\{q\in \mathcal P(Y):(x,q)\in A\}. $$
If I am not mistaken, we obtain $A' = (\mathrm{id}_X\times \pi_*)^{-1}(A)$, hence the question can be reduced to the measurability of the map $\pi_*$.
I think, a more general result holds true. Let $(\Omega,\mathcal F)$ and $(\Omega',\mathcal F')$ be arbitrary measurable spaces, and let $\mathcal P$ be a set of all probability measures on $(\Omega,\mathcal F)$ endowed with a $\sigma$-algebra $\mathcal A$ generated by evaluation maps $\theta_F:\mathcal P \to \Bbb R$ given by $\theta_F(p) = p(F)$ for any $F\in \mathcal F$ and $p\in \mathcal P$. Let $(\mathcal Pi',\mathcal A')$ be a corresponding measurable space of probability measures for $(\Omega',\mathcal F')$ and denote by $\theta'_{F'}$ the corresponding evaluation maps. For any measurable $\varphi:\Omega\to\Omega'$ it holds that $\varphi_*:\mathcal P\to\mathcal P'$ is measurable.
Proof: note that for any $p\in \mathcal P$ and $F'\in \mathcal F'$ it holds that $$ \theta'_{F'}(\varphi_*p) = p(\varphi^{-1}(F')) = \theta_{\varphi^{-1}(F')}(p) $$ hence $\theta'_{F'}\circ \varphi_* = \theta_{\varphi^{-1}(F')}$. Since $\mathcal A'$ is generated by evaluation maps, for the measurability of $\varphi_*$ it is necessary and sufficient that $\varphi_*^{-1}((\theta'_{F'})^{-1}(B))\in \mathcal A$ for any Borel $B\subseteq \Bbb R$. The latter fact is true since $$ \varphi_*^{-1}((\theta'_{F'})^{-1}(B)) = (\theta'_{F'}\circ \varphi_*)^{-1}(B) = (\theta_{\varphi^{-1}(F')})^{-1}(B)\in \mathcal A $$ since $\varphi^{-1}(F')\in \mathcal F$ and $\mathcal A$ is generated by maps $\theta_{F}$ with $F\in \mathcal F$.
In this case, you can say even more: under compatible Polish topologies, $\pi_*$ is not only measurable but continuous.
Indeed, suppose $U,V$ are Polish spaces and $F : U \to V$ is continuous. Then $F_* : \mathcal{P}(U) \to \mathcal{P}(V)$ is continuous in the weak topologies. The proof is immediate: suppose $\mu_n, \mu \in \mathcal{P}(U)$ with $\mu_n \to \mu$ weakly. Let $g : V \to \mathbb{R}$ be bounded and continuous. Then $g \circ F$ is a bounded continuous function on $U$, so we have $$\int_V g\,dF_* \mu_n = \int_U g \circ F \,d\mu_n \to \int_U g \circ F\,d\mu = \int_V g\,dF_*\mu.$$
For your example, fix compatible Polish topologies on $Y,Y'$, and set $U = Y \times Y'$, $V = Y$, and $F = \pi$, the projection map.