A problem from Isaacs's Finite Group Theory I was revisiting group theory in detail and reading Isaacs's Finite Group Theory book in my own time. Sorry that I am asking an exercise question but this is the one I am stuck completely. Any help will be really appreciated.
The problem is 1F.3 on page 40.

Let $G=NP$ be a finite group, where $N$ is a normal subgroup of $G$, and $P$ is a Sylow $p$-subgroup of $G$ with $N\cap P=1$, and assume that the conjugation action of $P$ on $N$ is faithful. Show that $P$ acts faithfully on at least one orbit of this action.

The given hint is that we have to consider $x\in N$ with the properly that $P\cap P^x$ is least in size and then to show that $P$ acts faithfully on the $P$-orbit containing $x$. My guess is that somehow we have to use Theorem 1.38 of the book but I can not figure out how. I have given Theorem 1.38 below; here $O_p(G)$ stands for the $p$-core of $G$ that is the unique largest normal $p$-subgroup of $G$, and it can be found by taking the intersection of all of the Sylow $p$-subgroups of $G$.

Thanks in advance.
 A: I think it is time to answer this question!
As has been pointed out in comments, we have $[O_p(G),N] \le N \cap O_p(G) = 1$ and $O_p(G) \le P$ so, since we are told that $P$ acts faithfully by conjugation on $N$, we must have $O_p(G)=1$.
Following the hint, choose $x \in N$ such that $|P \cap P^x|$ is minimal, and let $Q$ be the kernel of the action of $P$ on the orbit of $P$ on $N$ that contains $x$. So $Q \le P \cap P^x$ and clearly $Q \unlhd P$.
If we could prove that $Q \unlhd P^x$ then since we know from Theorem 1.38 of Isaacs that $O_p(G)$ is the largest subgroup of $P \cap P^x$ that is normal in both $P$ and $P^x$, we would have $Q \le O_p(G) =1$, so $P$ acts faithfully on this orbit, adn we are done.
So it remains to prove that $Q \unlhd P^x$ (which I found the most challenging part of this question).
Now $Q \le P^x \Rightarrow xQx^{-1} \le P$ so, for all $q \in Q$, we have $xqx^{-1}q^{-1} \le P$. But, since $N$ is normal in $G$, we also have $xqx^{-1}q^{-1} \le N$, so $xqx^{-1}q^{-1}=1$, and $x$ centralizes $Q$.
So $Q \unlhd P \Rightarrow Q = Q^x \unlhd P^x$, and we are done.
