Marginal distribution of random variable with multinomial sampling distribution and parameters $(n,\boldsymbol{p})$, where $n \sim $ Poisson Suppose you have:
$X|N \sim \text{MN}(N,p_1,p_2...p_J)$
$N \sim \text{Poisson}(\lambda)$
Where is the marginal distribution of $X$?
 A: Just to formalise what jbowman is telling you in the comments, for any valid argument vector $\boldsymbol{x}$ with $\sum x_j = n$, application of the law-of-total-probability gives:
$$\begin{equation} \begin{aligned}
\mathbb{P}(\boldsymbol{X}=\boldsymbol{x}) 
&= \mathbb{P}(\boldsymbol{X}=\boldsymbol{x} | N=n) \mathbb{P}(N=n) \\[6pt]
&= \text{Mu}(\boldsymbol{x} | n,\boldsymbol{p}) \text{Pois}(n|\lambda) \\[6pt]
&= \frac{n!}{\prod x_j!} \Big( \prod_{j=1}^J p_j^{x_j} \Big) \frac{\lambda^n}{n!} \exp(-\lambda) \\[6pt]
&= \prod_{j=1}^J \frac{(\lambda p_j)^{x_j}}{x_j !} \exp(-\lambda p_j) \\[6pt]
&= \prod_{j=1}^J \text{Pois}(x_j | \lambda p_j). \\[6pt]
\end{aligned} \end{equation}$$
(Note that the first line of this working is the law-of-total-probability, but with the sum being taken over only one value of $n$ that is consistent with the vector $\boldsymbol{x}$.)  Hence, we can see that the elements of $\boldsymbol{X}$ are independent, with marginal distributions $x_j \sim \text{Pois}(\lambda p_j)$.  
