I'm looking at Theorem 1.1 here:
Let G be a finite group with abelian Sylow p-subgroups. Let x and y be two elements in a Sylow p-subgroup P. If x and y are G-conjugate then they are $N_G(P)$-conjugate.
http://web.mat.bham.ac.uk/D.A.Craven/docs/lectures/fusionsystems.pdf
It says in the proof that $P$ and $P^g$ are both Sylow p-subgroups of the centraliser of y, but I'm not sure how this follows. I'd appreciate some help getting from the previous step to here.
There is a typo in the proof. It should say $y = x^g$, i.e. $y = gxg^{-1}$, rather than $x = y^g$.
If $P$ is a Sylow $p$-subgroup of a finite group $G$, and $H$ is another subgroup of $G$ such that $P \subseteq H$, then by definition, $P$ is also a Sylow $p$-subgroup of $H$.
So in your case, all you have to do is show that $P$ and $g^{}Pg^{-1}$ are contained in the centralizer $C_G(y)$. Since $y \in P$, and $P$ is abelian, certainly every element of $P$ commutes with $y$, i.e. $P \subseteq C_G(y)$.
As for $g^{}Pg^{-1}$, we are given that $y = gxg^{-1}$. If $z \in gPg^{-1}$, then $z = gpg^{-1}$ for some $p \in P$. Then
$$zy = (gpg^{-1})(gxg^{-1}) = gpxg^{-1} = gxpg^{-1} = (gxg^{-1})(gpg^{-1}) = yz$$
which shows that $gPg^{-1} \subseteq C_G(y)$ as well.