The answer is no. Indeed, for natural $n$ let
$$p_n(x):=1+\tfrac12\,\text{sign}\,\sin(2\pi nx)$$
if $x\in[0,1]$, with $p_n(x):=0$ if $x\notin[0,1]$. Then $p_n$ is a pdf, with the corresponding cdf $F_n$, so that
$$F_n(x)=\int_{-\infty}^x p_n(t)\,dt$$
for all real $x$. The cdf $F_n$ is continuously and strictly increasing on $[0,1]$, with $F_n(0)=0$ and $F_n(1)=1$. Hence, for each $u\in(0,1)$ there is a unique root $x=F_n^{-1}(u)$ of the equation $F_n(x)=u$. Moreover, the function $F_n^{-1}\colon(0,1)\to\mathbb R$ is continuous.

Further, $F_n(x)\to x$ for all $x\in[0,1]$ and hence $F_n^{-1}(u)\to u$ for all $u\in(0,1)$ (as $n\to\infty$).

Let now $X_n:=F_n^{-1}(U)$ and $X:=U$, where $U$ is any random variable uniformly distributed on $(0,1)$. Then $X_n\to X$ almost surely. Also, $F_n$ is the cdf of $X_n$ and hence $p_n$ is the pdf of $X_n$.

However, the pdf $p_n$ of $X_n$ does not converge to the pdf (say $p$) of $X$ even in measure -- because otherwise, by dominated convergence, $p_n$ would converge to $p$ in $L^1$, whereas the $L^1$ norm of $p_n-p$ is
$$\int_0^1|p_n(x)-1|\,dx=\frac12\not\to0.$$

On a positive note, the almost sure convergence of $X_n$ to $X$ will of course imply the convergence of $X_n$ to $X$ in distribution, so that the pdf $p_n$ of $X_n$ will converge to the pdf $p$ of $X$ in the weak sense that
$$\int_{\mathbb R}f(x)p_n(x)\,dx\to\int_{\mathbb R}f(x)p(x)\,dx$$
for each bounded continuous function $f\colon\mathbb R\to\mathbb R$.