Poisson conditional sum

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Let $N \sim Poisson(\lambda)$ r.v. independent $\{X_k, k = 1, 2, \ldots\}$, such that are independent with each other and $P(X_k = 1) = 1 - P(X_k = 0) = p \in (0, 1)$. Let define $K =\sum_{i=1}^{N} X_i$. Calculate

$P[K=k|N=n]$, and

$P[N=n|K=k]$.

For the first part I did the following: I know that $\sum_{i=1}^{n}X_i \sim Bin(n,p)$

$$P[K=k|N=n]=P\left[\sum_{i=1}^{n}X_i=k\right]=\binom{n}{k}p^{k}(1-p)^{n-k}.$$

But I got stuck in the second exercise, I tried with Bayes theorem but I do not how to conclude. Is this right? Any help?