Combination of Poisson and binomial distribution

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.