$\sum_{n=0}^\infty \binom{n}{k}p^k(1-p)^{n-k}\frac{\lambda^ne^{-\lambda}}{n!} = \frac{(\lambda p)^ke^{-\lambda p}}{k!}$
That is, given a random variable $X$ with Poisson distribution $X \sim \operatorname{Poisson}(\lambda)$, and, given $X = n$, random variable $Y$ is distributed by a binomial such that $Y \sim B(n,p): p \in [0,1]$. Given $n$, $P(Y = k|X = n) = \binom{n}{k}p^k(1-p)^{n-k}$. Given $\lambda$, $P(X=n) = \dfrac{\lambda^ne^{-\lambda}}{n!}$. Therefore, $P(Y = k|\lambda) = \sum_{n=0}^\infty P(Y = k | X = n)P(X = n | \lambda) = \sum_{n=0}^\infty \binom{n}{k}p^k(1-p)^{n-k}\dfrac{\lambda^ne^{-\lambda}}{n!}$. What properties and identities can be used to demonstrate the equality above?