Combination of Poisson and binomial distribution I'm working on the following problem

Each time you flip a certain coin, heads appears with probability $p$. Suppose that you flip the coin a random number of $N$ times, where $N$ has the Poisson distribution with parameter $\lambda$ and is independent of the outcomes of the flips. Find the distributions of the numbers $X$ and $Y$ of the resulting heads and tails, respectively, and show that $X$ and $Y$ are independent. 

What I tried, is conditioning on the value of $N$:
\begin{eqnarray}
\mathbb{P}(X=x) & = & \sum_{k=0}^{\infty}\mathbb{P}(X=x \ | \  N=k)\mathbb{P}(N=k)\\
& = & \sum_{k=0}^{\infty}\binom{k}{x}p^x(1-p)^{k-x}\frac{\lambda^ke^{-\lambda}}{k!}\\
& = & \sum_{k=x}^{\infty}\binom{k}{x}p^x(1-p)^{k-x}\frac{\lambda^ke^{-\lambda}}{k!}.\\
\end{eqnarray}
Similarly, for $Y$ i found $$\mathbb{P}(y=y)=\sum_{k=y}^{\infty}\binom{k}{y}p^{k-y}(1-p)^y\frac{\lambda^ke^{-\lambda}}{k!}.$$
I tried to work this out but I didn't seem to go anywhere. The answer should be that $X \sim Pois(\lambda p)$ and because of symmetry we would have $Y \sim Pois(\lambda (1-p))$.
Can anyone provide some help about how to from where I came to $X \sim Pois(\lambda p)$? Thanks in advance.
 A: I would suggest to use moment-generating functions (MGF): simpler, faster proof. Namely, you have, for $t\in\mathbb{R}$,
$$\begin{align}
\mathbb{E} e^{tX}
&= \mathbb{E}[ \mathbb{E}[ e^{tX} \mid N ] ]
\stackrel{(\dagger)}{=} \mathbb{E}[ (1-p+pe^{t})^N ]\\
&= \mathbb{E}[ e^{N\ln(1-p+pe^{t})} ]
\stackrel{(\ddagger)}{=} \exp(\lambda(e^{\ln(1-p+pe^{t})}-1))\\
&= \exp(\lambda((1-p+pe^{t})-1))\\
&= \exp(\lambda p(e^{t}-1))
\end{align}$$
where $(\dagger)$ uses the expression of the MGF of a Binomial distribution with parameters $N$ and $p$, and $(\ddagger)$ that of the MGF of a Poisson  distribution with parameter $\lambda$ (applied to the argument $t'\stackrel{\rm def}{=}\ln(1-p+pe^{t})$).
At the end, you get that, for every $t\in\mathbb{R}$,
$$
\mathbb{E} e^{tX} = \exp(\lambda p(e^{t}-1)) \tag{$\ast$}
$$ 
which is the MGF of a Poisson distribution with parameter $\lambda p$. As the MGF characterizes the distribution (when it exists), we have the result.

However, if you want to finish your computation: here how it goes. I assume $p\neq 1$, otherwise the answer is trivial.
\begin{align}
\mathbb{P}\{X=n\} &= \sum_{k=n}^\infty \binom{k}{n}p^n(1-p)^{k-n} \frac{\lambda^k e^{-\lambda}}{k!}\\
&= e^{-\lambda}\frac{p^n}{(1-p)^n}\sum_{k=n}^\infty \binom{k}{n}(1-p)^{k} \frac{\lambda^k}{k!}\\
&= e^{-\lambda}\frac{p^n}{(1-p)^n}\sum_{k=n}^\infty \frac{k!}{n!(k-n)!}(1-p)^{k} \frac{\lambda^k}{k!}\\
&= e^{-\lambda}\frac{p^n}{n!(1-p)^n}\sum_{k=n}^\infty \frac{1}{(k-n)!}(1-p)^{k} \lambda^k\\
&= e^{-\lambda}\frac{p^n}{n!(1-p)^n}\sum_{\ell=0}^\infty \frac{1}{\ell!}(1-p)^{\ell+n} \lambda^{\ell+n}\\
&= e^{-\lambda}\frac{(\lambda p)^n}{n!}\sum_{\ell=0}^\infty \frac{(1-p)^{\ell} \lambda^{\ell}}{\ell!}\\
&= e^{-\lambda}\frac{(\lambda p)^n}{n!}e^{\lambda(1-p)}
= \boxed{e^{-\lambda p}\frac{(\lambda p)^n}{n!}}
\end{align}
and you get the probability mass function of a Poisson r.v. with parameter $\lambda p$, as desired.
