Given $H \le G$ and $P \in \text{Syl}_{p}(G)$, to show that $\exists a \in G, aPa^{-1} \cap H \in \text{Syl}_{p}(H)$. The problem is as follows:

$G$ is a finite group. $H \le G$ and $P$ is a Sylow $p$-subgroup of $G$. 
  To prove that: $\exists a \in G$ such that $aPa^{-1} \cap H$ is a Sylow $p$-subgroup of $H$.

My trail is as follows:
We have $aPa^{-1} \cap H \le H$. To show that $aPa^{-1} \cap H$ is a Sylow $p$-subgroup of $H$, it is sufficient to prove that $p \nmid [H : aPa^{-1} \cap H]$.
Because $|H (aPa^{-1})| = \frac{|H| \cdot |aPa^{-1}|}{|aPa^{-1} \cap H|}$, we get $[H : aPa^{-1} \cap H] = [H(aPa^{-1}) : aPa^{-1}]$.
If we can show that $H(aPa^{-1}) \le G$, then we have $[H : aPa^{-1} \cap H] = \frac{|H(aPa^{-1})|}{|aPa^{-1}|} = \frac{|H(aPa^{-1})|}{|P|} \mid \frac{|G|}{|P|}$.
Therefore, $p \nmid [H : aPa^{-1} \cap H]$ since $p \nmid \frac{|G|}{|P|}$.
However, can we show that $\exists a \in G$ such that $H(aPa^{-1}) \le G$?
Or, if the argument is not correct, could you please offer me some hints?

Let $Q$ be a Sylow $p$-subgroup of $H$. It is a $p$-subgroup of $G$ and hence contained in a Sylow $p$-subgroup of $G$ and this is conjugate to $P$.
