Let $(Z_k)_{k\in\mathbb N}$ be a Markov chain on $(\Omega,\mathcal F,\mathbb P)$ taking values in $(\mathbb N,2^{\mathbb N})$. Let $p\in [0,1]$ and define $F:\mathbb N\times 2^\mathbb{N}\to [0,1]$ such that $F(n,B)$ being the law of negative binomial random variable with parameter $(n,p)$, i.e. \begin{align*} F(n,B):=\sum_{k\in B} \binom{n+k-1}{k}p^n (1-p)^k \end{align*} Now assume that we have \begin{align*}\tag{$\triangle$} \mathbb P(Z_{k+1}\in B\mid Z_k) = F(Z_k,B) \ \ \ \text{ a.s. } \end{align*}
Question. Does there exist a sequence of i.i.d. random variables $(X_{k, i} )_{k, i\in\mathbb N}$ on the same space as $Z_k$ such that $X_{k, i} $ is geometric distributed with parameter $p$ and $Z_{k+1}$ can be written as $$Z_{k+1}=\sum_{i=1}^{Z_k} X_{k, i} \tag{ $*$} $$
A little bit of motivation and context.
This question arose in a question where I had to show that $(Z_k)_{k\in\mathbb N}$ is a Branching process. I actually know how to solve that particular question in another way (there is a story behind it). However, for me it was easier to reduce the problem in showing $(*)$ given I know that $(\triangle)$ holds.
It is actually not so crazy to believe $(*)$, because then we have $(\triangle)$. The fact that $(Z_k)_{k\in\mathbb N}$ is a Markov chain also makes me believe it. I also know that given a distribution, then there exists a random variable with that distribution. But I don't know how to step that up with conditional distribution... I think that the result would not hold if $(Z_k)_{k\in\mathbb N}$ was not a Markov chain.
I appreciate if someone lets me know whether this is true or a reference for a proof of it.