$\newcommand{\ep}{\varepsilon}
\newcommand{\de}{\delta}
\newcommand{\B}{\mathcal B}
\newcommand{\K}{\mathcal K}
\newcommand{\NN}{\mathcal N}
\newcommand{\PP}{\mathcal P}
\newcommand{\supp}{\operatorname{supp}}$

The answer to your question is yes.

Indeed, recall that the weak$^{\star}$ topology on $\PP(Y)$ is metrized by the Lévy--Prokhorov metric $L$ defined by the condition: for any real $h>0$ and any $P,Q$ in $\PP(Y)$,
\begin{equation}
L(P,Q)\le h\iff\forall B\in\B(Y)\ P(B)\le Q(B_h)+h,
\end{equation}
where $\B(Y)$ is the Borel $\sigma$-algebra over $Y$ and $B_h$ is the $h$-neighborhood of $B$.

Let $\supp P$ denote the support of $P$. Let $\PP(K):=\{Q\in\PP(Y)\colon\supp Q\subseteq K\}$.

**Lemma 1:** Take any compact $\K\subseteq\PP(Y)$ and any real $\ep>0$. Then there is a compact $K\subseteq Y$ such that for each $P\in\K$ there is some $Q_P\in\PP(K)$ such that
\begin{equation}
\sup_{P\in\K}L(P,Q_P)\le\ep.
\end{equation}

*Proof:* By Prokhorov's theorem, for each real $\de\in(0,1)$ there is a compact $K\subseteq Y$ such that for each $P\in\K$ we have $P(K)\ge1-\de$. For all $B\in\B(Y)$, let then
\begin{equation}
Q_P(B):=\frac{P(B\cap K)}{P(K)}.
\end{equation}
Then for all $P\in\K$ and all $B\in\B(Y)$ we have
\begin{equation}
Q_P(B)\le\frac{P(B)}{1-\de}\le P(B)+\frac\de{1-\de}=P(B)+\ep
\end{equation}
if $\de=\ep/(1+\ep)$. So, indeed $L(P,Q_P)\le\ep$ for all $P\in\K$. $\Box$

Let $\PP_{fin}(Y)$ denote the set of all $Q\in\PP(Y)$ with a finite support.

**Lemma 2:** Take any compact $\K\subseteq\PP(Y)$ and any real $\ep>0$. Then for each $P\in\K$ there is some $R_P\in\PP_{fin}(Y)$ such that
\begin{equation}
\sup_{P\in\K}L(P,R_P)\le2\ep.
\end{equation}

*Proof:* Let $K\subseteq Y$ and $Q_P\in\PP(K)$ for $P\in\K$ be as in Lemma 1. Since $K$ is compact, there is a finite Borel-measurable partition $(C_{\ep;j})$ of $K$ such that for each $j$ there is a point $y_j\in Y$ such that $C_{\ep;j}$ is contained in the ball $B_\ep(y_j)$ of radius $\ep$ centered at $y_j$.

For any $P\in\PP(Y)$ and any $B\in\B(Y)$, let
\begin{equation}
Q_P^\ep(B):=\sum_j Q_P(C_{\ep;j})1_{y_j\in B}=Q_P(C_{\ep;B}),
\end{equation}
where
\begin{equation}
C_{\ep;B}:=\bigcup_{j\colon y_j\in B}C_{\ep;j}.
\end{equation}
Then $Q_P^\ep\in\PP_{fin}(Y)$. Moreover,
$$L(Q_P^\ep,Q_P)\le\ep$$
for all $P\in\PP(Y)$, because $C_{\ep;B}\subseteq B_\ep$ for all $B\in\B(Y)$. Letting now $R_P:=Q_P^\ep$, we see that Lemma 2 follows by Lemma 1. $\Box$

Finally, take any function $P_\cdot\in C(X,\PP(Y))$, which is a continuous map $X\ni x\mapsto P_x\in\PP(Y)$. Take also any compact $K_X\subseteq X$. Then the image $\K:=\K_{K_X}:=P_\cdot(K_X)$ of the compact $K_X$ under the continuous map $P_\cdot$ is compact. So, by Lemma 2, for each real $\ep>0$ and each $x\in K_X$ there is some $R_{\ep;x}\in\PP_{fin}(Y)$ such that
\begin{equation}
\sup_{x\in K_X}L(P_x,R_{\ep;x})\le2\ep.
\end{equation}
This means that indeed $C(X,\PP_{fin}(Y))$ is dense in $C(X,\PP(Y))$, in the required sense.