My actual question will appear at the bottom of this posting.

Suppose $$ \Pr(\Lambda\in d\ell) = \frac 1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m}\, \left( \frac{d\ell} m \right) \text{ for } \ell>0, $$ so $\Lambda$ is a random variable with a gamma distribution with expected value $\alpha m$ and variance $\alpha m^2$.

Let the conditional distribution of the random variable $N$, given $\Lambda$, be $$ N\mid\Lambda \sim \operatorname{Poisson}(\Lambda). \tag 1 $$ Then the marginal (``unconditional'') distribution of $N$ is a negative binomial distribution: \begin{align*} & \Pr(N=n) \\[10pt] = {} & \operatorname E(\Pr(N=n\mid\Lambda)) = \operatorname E\left( \frac{\Lambda^n e^{-\Lambda}}{n!} \right) \\[10pt] = {} & \int_0^\infty \frac{\ell^n e^{-\ell}}{n!} \cdot \frac1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m} \, \left( \frac{d\ell} m \right) \\[8pt] = {} & \frac{(n+\alpha-1)(n+\alpha-2)(n+\alpha-3)\cdots\alpha}{n!} \left( \frac m{m+1} \right)^n \left( \frac1{m+1} \right)^\alpha \\[8pt] = {} & \binom{-\alpha}{\phantom{-}n} p^\alpha (-q)^n \text{ for } n\in\{0,1,2,3,\ldots\} \tag 2 \end{align*} (so that $p+q=1$), where $$ \binom{-\alpha}{\phantom{-}n} = \frac{\overbrace{-\alpha(-\alpha-1)(-\alpha-2)\cdots(-\alpha-n+1)}^\text{$n$ factors}}{n!}. $$ This has expected value $\alpha q/p$ and variance $\alpha q/p^2$.

Perhaps it is less widely known that the same negative binomial distribution arises as a compound Poisson distribution:

Suppose $\Pr(X=x) = \dfrac{-q^x}{x\log(1-q)}$ for $x=1,2,3,\ldots$, and let $X_1,X_2,X_3,\ldots$ be independent copies of this random variable. (This is called the logarithmic series distribution since $\sum_{x=1}^\infty q^x/x = -\log(1-q).$) Suppose $M\sim\operatorname{Poisson}(-\alpha\log(1-q))$.

Then $$ N = \sum_{i=1}^M X_i \tag 3 $$ also has the same negative binomial distribution that appears on line $(2)$ above.

My question is: How can we construct a single probability space that is the domain of all of the random variables mentioned here, in such a way that the $N$ defined on line $(1)$ above and the $N$ defined on line $(3)$ above are not just two random variables sharing the same distribution, but are just one and the same random variable?