= Solution
Take an <independent and identically distributed> sequence with \b[$X_i\sim\operatorname{Bernoulli}(r)$], independent of $N$. Conditional on $N=n$, the <random sum> $\sum_{i=1}^nX_i$ has <binomial distribution> with parameters $n,r$, including the <empty sum> at $n=0$. This construction therefore has exactly the prescribed joint law of $(N,Z)$ and represents $Z$ as the required <random sum>.
For a literal representation that retains an already given pair $(N,Z)$, enlarge the <probability space> by auxiliary randomness. Conditional on $N=n,Z=z$, choose uniformly a size-$z$ subset of $\{1,\ldots,n\}$, put ones at those indices and zeros at the others, and append independent <Bernoulli> variables of parameter $r$. Given $N=n$, a specified length-$n$ pattern with $z$ ones has probability $\binom nzr^z(1-r)^{n-z}/\binom nz=r^z(1-r)^{n-z}$. Thus the entire sequence is <independent and identically distributed> and independent of $N$, while its first $N$ terms sum to the original $Z$.
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