Let $X\sim \operatorname{Poisson}(\lambda)$ and $Y$ a random variable in $\mathbb{N}$. The joint probability function is $$P(Y=k,X=n)=e^{-\lambda}\frac{(\lambda p)^k(\lambda (1-p))^{n-k}}{k!(n-k)!}$$ for $k\in \{0,\dots,n\}$ otherwise $0$.
Show that $Y\sim \operatorname{Poisson}(p\lambda)$.
I know from a previous exercise that $P(Y=k\mid X=n)\sim B(n,p)$ so $$P(Y=k,X=n)=e^{-\lambda}\frac{(\lambda p)^k(\lambda (1-p))^{n-k}}{k!(n-k)!}=\binom{n}{k}p^k(1-p)^{n-k}\frac{e^{-\lambda}\lambda^n}{n!}$$ and when I put $P(Y=n,X=n)$ I get the Poisson distribution with parameter $\lambda p$ but this is only for the case $P(Y=X)$. Unfortunately I don't know how to compute $$P(Y=k)=\sum\limits_{n=1}^\infty e^{-\lambda}\frac{(\lambda p)^k(\lambda (1-p))^{n-k}}{k!(n-k)!}$$
I also tried to work with Bayes theorem, but since the next exercise is to compute $P(Y=k\mid X=n)$ I think this is not the appropriate way.
$$ \Pr(Y=k,X=n)=e^{-\lambda}\frac{(\lambda p)^k(\lambda (1-p))^{n-k}}{k!(n-k)!} \text{ for } 0\le k\le n $$ So \begin{align} & \Pr(Y=k) \\[10pt] = {} & \Pr\Big( (Y=k\ \&\ X=k) \text{ or } (Y=k\ \&\ X=k+1) \text{ or } (Y=k\ \&\ X=k+2) \text{ or } \cdots \Big) \\[10pt] = {} & \sum_{n=k}^\infty \Pr(Y=k\ \&\ X=n) = \sum_{n=k}^\infty e^{-\lambda} \frac{(\lambda p)^k(\lambda (1-p))^{n-k}}{k!(n-k)!} \\[10pt] = {} & \frac{e^{-\lambda}}{k!}(\lambda p)^k \sum_{n=k}^\infty \frac{(\lambda(1-p))^{n-k}}{(n-k)!}. \end{align} The last step is justified by the fact that the factors that were pulled out do not depend on $n$, which is the index of summation; they don't change as $n$ goes from $k$ to $\infty$.
As $n$ goes from $k$ to $\infty$, $m=n-k$ goes from $0$ to $\infty$, and the last expression becomes $$ \frac{e^{-\lambda}}{k!} (\lambda p)^k \sum_{m=0}^\infty \frac{(\lambda(1-p))^m}{m!}. $$ The last sum is $\displaystyle \sum_{m=0}^\infty \frac{x^m}{m!} = e^x$ for $x=\lambda(1-p)$.