If $P\le G$ is a sylow-$p$ and $Q$ is any $p$ subgroup, then $Q\cap P = Q\cap N(P).$ 
If $P\le G$ is a sylow-$p$ and $Q$ is any $p$ subgroup, then $Q\cap P = Q\cap N(P).$

I'd appreciate any help.
I have a proof from some old notes but it says that it is sufficient to prove that if $qPq^{-1}=P$, and $q\in Q,$ $q\in P$, then we have proven the proposition. This doesn't seem true.
Maybe I'm wrong.
 A: Yet another proof: $Q \cap N_G(P)$ is a $p$-subgroup of $N_G(P)$. Now apply Sylow Theory in $N_G(P)$, telling you that $Q \cap N_G(P)$ must be contained in some Sylow $p$-subgroup of $N_G(P)$. Since $P  \unlhd N_G(P)$, this Sylow $p$-subgroup must be $P$, whence $Q \cap N_G(P) \subseteq P$. 
A: I think that easy to see $Q\cap P\subseteq Q\cap N(P).$ 
Now assume $|Q|=p^{k_1},|P|=p^{k_2},|N(P)|=p^{k_2}n_2$ and $|QN(P)|=p^{k_2}n_3$ where $k_1,k_2,n_2,n_3\in\mathbb{N}$ and $k_2\geq k_1,n_3\geq n_2\geq 1$ and $n_2,n_3$ not divisible by $p.$ 
We have $$|Q\cap N(P)|=\dfrac{|Q|\cdot|N(P)|}{|QN(P)|}=\dfrac{p^{k_1}\cdot p^{k_2}n_2}{p^{k_2}n_3}=p^{k_1}\dfrac{n_2}{n_3}.$$ Thus, $n_3\mid n_2\Rightarrow n_3\leq n_2\Rightarrow n_2=n_3.$ Therefore $$|Q\cap N(P)|=p^{k_1}.$$ 
By the same way, we have $|Q\cap P|=p^{k_1}\Rightarrow |Q\cap P|=|Q\cap N(P)|.$ Thus $Q\cap P=Q\cap N(P).$
A: It is enough to show that $Q\cap N_G(P) \subseteq Q\cap P$, since $P\subseteq N_G(P)$. Say $q\in Q\cap N_G(P)$. Then $\langle P,q\rangle\subseteq N_G(P)$, so $P\triangleleft \langle P,q\rangle$.
Now, $\langle P,q\rangle/P$ is generated by the image of $q$, and so has order a power of $p$; but its order is also prime to $p$ since you are moding out by a Sylow $p$-subgroup. Therefore, the order is $p^0$. That is, $P=\langle P,q\rangle$, so $q\in P$, which is what you wanted to prove. 

What your notes likely mean is that if you take $q\in Q$, with $qPq^{-1}=P$, (that is, $q\in Q\cap N_G(P)$), you are supposed to deduce that $q\in P$. That is, it just tells you what you need to show.
