$\newcommand{\XXX}{\mathbb X}\newcommand{\de}{\delta}$The answer is yes.
Indeed, for any nonnegative measurable function $f$ defined on $\XXX$, letting $Nf:=\int f\,dN=\sum_{i=1}^M f(X_i)$ and assuming that $M$ is independent of the $X_i$'s, we have
\begin{equation*}
Ee^{-Nf}=\sum_{m\ge0}q_m(Ee^{-f(X)})^m, \tag{1}\label{1}
\end{equation*}
where $X:=X_1$ and $q_m:=P(M=m)$.
Next, we may write
\begin{equation*}
N_p=\sum_{i=1}^M \de_{X_i}\,1(U_i<p(X_i)),
\end{equation*}
where the $U_i$'s are independent random variables (r.v.'s) uniformly distributed on $[0,1]$ and independent of the $X_i$'s and $M$.
So, letting $U:=U_1$, we have
\begin{equation*}
Ee^{-N_pf}=\sum_{m\ge0}q_m(Ee^{-f(X)1(U<p(X))})^m.
\end{equation*}
Next,
\begin{equation*}
\begin{aligned}
&Ee^{-f(X)1(U<p(X))} \\
&=\int P(X\in dx)\int_0^{p(x)}du\,e^{-f(x)}+\int P(X\in dx)\int_{p(x)}^1 du \\
&=\int P(X\in dx)[p(x)e^{-f(x)}+1-p(x)] \\
&=1+\int P(X\in dx)p(x)(e^{-f(x)}-1).
\end{aligned}
\end{equation*}
Introducing now a random element $Z$ in $\XXX$ with the distribution
\begin{equation*}
P(Z\in dx)=\frac{P(X\in dx)p(x)}c,\quad\text{with}\quad c:=\int P(X\in dy)p(y)\in(0,1]
\end{equation*}
(and avoiding the trivial case with $c=0$), we get
\begin{equation*}
Ee^{-f(X)1(U<p(X))}=1+c\int P(Z\in dx)(e^{-f(x)}-1) \\
=cEe^{-f(Z)}+1-c.
\end{equation*}
So,
\begin{equation*}
(Ee^{-f(X)1(U<p(X))})^m=\sum_{j=0}^m\binom mj c^j(1-c)^{m-j}(Ee^{-f(Z)})^j
\end{equation*}
and hence
\begin{equation*}
\begin{aligned}
Ee^{-N_pf}&=\sum_{m\ge0}q_m(Ee^{-f(X)1(U<p(X))})^m \\
&=\sum_{m\ge0}q_m \sum_{j=0}^m\binom mj c^j(1-c)^{m-j}(Ee^{-f(Z)})^j \\
&=\sum_{j\ge0}r_j(Ee^{-f(Z)})^j,
\end{aligned}
\end{equation*}
where $r_j:=\sum_{m=j}^\infty q_m\binom mj c^j(1-c)^{m-j}$.
So,
\begin{equation*}
Ee^{-N_pf}=\sum_{j\ge0}r_j(Ee^{-f(Z)})^j. \tag{2}\label{2}
\end{equation*}
Choosing $f=0$ in \eqref{2}, we see that $\sum_{j\ge0}r_j=1$.
Finally, comparing \eqref{2} with \eqref{1}, we conclude that any "thinned" mixed binomial process is indeed a "nonthinned" mixed binomial process: $N_p$
is equal in distribution to the random measure
\begin{equation}
\tilde N_p:=\sum_{i=1}^J \de_{Z_i},
\end{equation}
where the $Z_i$'s are independent copies of $Z$ and $J$ is a r.v. independent of the $Z_i$'s and such that $P(J=j)=r_j$ for all $j$. $\quad\Box$
The probabilistic meaning of the random measure $\tilde N_p$ is transparent: Refer to $X_i$ as the random position of the $i$th particle in the space $\XXX$. Then, for each $i$, the r.v. $p(X_i)$ is the conditional probability that the $i$th particle will survive (be retained) given its position $X_i$. So, $c=\int P(X\in dy)p(y)=Ep(X_i)$ is, for each $i$, the "total", space-averaged survival probability of the $i$th particle. So, the distribution of each of the independent $Z_i$'s is the conditional distribution of the position $X_i$ of the $i$th particle given its survival. Finally, the conditional distribution of the random time extent $J$ given the random time extent $M$ is the binomial distribution of the number of successes in $M$ independent trials with success probability $c$ in each trial, if success is defined as survival.
