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?
Answer my own question:
Define a group action: $H$ acts on $G/P$ by left multiplication, generating $k$ orbits $O_1, O_2, \cdots, O_k$.
$|G/P| = \sum_{i=1}^{i=k} |O_i|$. Because $p \nmid |G/P|$, there exists $aP \in G/P$ such that $p \nmid |O_{aP}| = [H : \text{Stab}(aP)]$. Thus, $\text{Stab}(aP) \in \text{Syl}_{p}(H)$.
On the other hand, $\text{Stab}(ap) = \{ h \in H: haP = aP \} = \{ h \in H: a^{-1}ha \in P \} = \{ h \in H : h \in aPa^{-1} \} = aPa^{-1} \cap H.$