Let $\Theta$ be a parameter space, which is a complete, separable metric space endowed with a Borel $\sigma-algebra$ $\mathcal{B}(\Theta)$.
Then for each $\theta \in \Theta$, $P_{\theta}$ is a probability measure on the measurable space $(X,\mathcal{A})$, such that for each $A\in\mathcal{A}$, $\theta \mapsto P_{\theta}(A)$ is $\mathcal{B}(\Theta)$ measurable.
Does the measurability of $P_{\theta}$ mean the following ??
We know that for fixed $\theta$, $P_{\theta}:\mathcal{A}\rightarrow [0,1]$, now if we view the probability function $P_{\theta}$ as function of $\theta$ for a fixed set $A\in \mathcal{A}$, we will have $P_{\theta}(A):\mathcal{B}(\Theta)\rightarrow [0,1]$??
Then the $\mathcal{B}(\Theta)$ measurability will mean the following ??
For each $A\in \mathcal{A}$, $P_{\theta}(A)$ is measurable in $\mathcal{B}(\Theta)$,
if for every $E\in\mathcal{B}([0,1])$ the $\left \{ \theta \in \Theta:P_{\theta}(A)\in E \right \}\subset\mathcal{B}(\Theta)$
The definition means that for each fixed $A\in \mathcal{A}$ the map
$$ f_A:\Theta \to [0,1],\quad \theta \mapsto P_\theta (A) $$
is $\mathcal{B}(\Theta )$-measurable. Note that each $P_\theta $ is a measure, not a function, so it doesn't make sense to ask about if it is measurable.