Third part of Sylow's theorem [from Herstein's book] I was working on Third Sylow's theorem from Herstein's book. 
1) Can anyone explain why the sum $\sum \limits_{x\in N(P)}o(Px)$ is equal to $o(N(P))$? I have tried some ways but no results.
2) Since $p^{n+1}\nmid o(N(P))$ and $\dfrac{p^{n+1}u}{o(N(P))}$ is an integer why $\dfrac{p^{n+1}u}{o(N(P))}$ must be divisible by $p$? 
Would be grateful if somebody will provide detailed explanations to my questions.
 A: In the sum $\sum_{x\in N(p)}o(Px)$, $o(Px)=o(P)$ and so $$\sum_{x\in N(p)}o(Px)=\sum_{x\in N(p)}o(P)=o(P)\sum_{x\in N(P)}1.$$ Now the sum $\sum_{x\in N(P)}1$ is the number of elements $x$ in $N(P)$ for which the cosets $Px$ are distinct. Since $P$ is a subgroup of $N(P)$, this is just the index of $P$ in $N(P)$ i.e., $o(N(P))/o(P)$ and so the result follows.

Edit: To see why the cosets are distinct, remember that the double cosets $PxP$ form equivalence classes in $G$ so $o(G)=\sum o(PxP)$ where the sum extends over all $x$ for which $PxP$ are distinct. But the corresponding cosets $Px$ are subsets of the double cosets $PxP$ and therefore if the double cosets are distinct, then so are the "single" cosets. So in the sum $\sum_{x\in N(P)}1$, $x$ actually has two properties: $x\in N(P)$ and the cosets $Px$ are distinct.

As for the second question, since $p^{n+1}$ can't divide $o(N(P))$, so the highest power of $p$ that $o(N(P))$ can have is $n$ and therefore $p^{n+1}u/o(N(P))$ has at least one $p$ as a factor.
