Q:

$\exists g\in G: H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$

Question:
If $P$ is a Sylow $p-$subgroup of $G$ and $H\leq G$ with $p||H|$ then $\exists g\in G: H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$.
Attempt:
We consider the action $H\times G/P\to G/P$ with $(h,xP)\to hxP$ and we know that 
$$|G/P|=\sum_{x\in S}|[xP]_{H}|=\sum_{x\in S}|H:H\cap xPx^{-1}|$$
since $|[xP]_{H}|=|H:Stab_H(xP)|=|H\cap xPx^{-1}|$.
Now, $P$ is a Sylow $p-$subgroup of $G$ so $p\nmid |G/P|=m$ so from above $\exists g\in G:p\nmid |H:H\cap gPg^{-1}|$

If $|H:H\cap gPg^{-1}|\not=1$: we have that $p\nmid |H:H\cap gPg^{-1}|$ and $H\cap gPg^{-1}$is a $p-$subgroup of $H$(please explain this statement) so $H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$.
If $|H:H\cap gPg^{-1}|=1$ then $H=H\cap gPg^{-1}\Rightarrow H\leq gPg^{-1}(= $$p-$group) and therefore $H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$.

Is this proof correct? Is there a quicker way to prove the initial statement?

A:

For the "please explain this statement" part: The conjugate $gPg^{-1}$ has the same order with $P$, hence it is a power of $p$. Therefore, $H\cap gPg^{-1}$, being a subgroup of $gPg^{-1}$, has order dividing $|gPg^{-1}|$, therefore, it a power of $p$. Hence, it is a group of order $p^k$ for some $k\neq 0$, hence a $p$-subgroup of $H$.
Now, in the first bullet, the only "rough" part is why $H\cap gPg^{-1}$ is a Sylow $p$-subgroup. That is because it a non-trivial $p$-subgroup with $p$ not dividing $[H:H\cap gPg^{-1}]$, hence a maximal $p$-subgroup (aka Sylow).
In the second bullet, it is clear that $H\cap gPg^{-1}=H$ hence it is trivially a $p$-Sylow subgroup.
Keep in mind, η Ταλέλλη δεν κάνει λάθος...

