Let $A \unlhd G$, then $n_p(G)=n_p(G/A)n_p(PA)$ for $P\in Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$ $n_p(G)=n_p(G/A)n_p(PA)$ for $P\in Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$
This result is from the paper Guo, W., & Vdovin, E. P. (2018). Number of Sylow subgroups in finite groups. Journal of Group Theory, 21(4), 695-712.
The proof is as follows:

Consider a natural homomorphism $\bar{}:G\rightarrow \overline{G}=G/A$. Clearly, $n_p(PA)$ is the number of all Sylow $p$-subgroups $Q$ of $G$ such that $\overline{P}=\overline{Q}$ since $\overline{P}=\overline{Q}$ if and only if $Q\leq PA$. By Sylow theorem, for every $Q\in Syl_p(G)$, there exists $x\in G$ such that $P^x=Q$, so $(PA)^x=QA$. This shows that $n_p(PA)$ does not depend on the choice of $P$, and so it is the same for any Sylow $p$-subgroups of $G$. Thus we obtain $n_p(G)=n_p(G/A)n_p(PA)$. By Sylow theorem, $$n_p(PA)=|PA:N_{PA}(P)|=|A:N_A(P)|$$ where the second identity follows from the Dedekind theorem.

Several questions/problems I have:
(i)If $Q\leq PA$, I can only show that $\overline{Q}\leq \overline{P}$. I can't see why $\overline{P}\leq \overline{Q}$ in this case.
(ii)I can't see why the previous argument would directly lead to the conclusion $n_p(G)=n_p(G/A)n_p(PA)$.
(iii)I search a lot about the Dedekind theorem mentioned in the text but the result I found is for continuity on real axis or for Galois group.
 A: Let me offer a slightly different proof, which might illuminate what's going on.
The Sylow subgroups of $G/A$ look like $PA/A$, where $P$ is a Sylow subgroup of $G$.  Thus $G$ acts on the Sylow subgroups of $G/A$, and it acts transitively (since it acts transitively on its own Sylow subgroups). This action is just the conjugation action on $PA$ up in $G$. By the orbit-stabilizer theorem, then,
$$ n_p(G/A) = [G:N_G(PA)] $$
We also have $n_p(G)=[G:N_G(P)]$.  Now if $g\in N_G(P)$, then $g\in N_G(PA)$, since $A$ is normal.  So we can write
$$ n_p(G) = n_p(G/A)\cdot[N_G(PA):N_G(P)]$$
That last number is just $n_p(N_G(PA))$. But $P\le PA\lhd N_G(PA)$, so every Sylow subgroup of $N_G(PA)$ is one of $PA$; thus that last number is also $n_p(PA)$, which gives you the first claim.
For your second claim, let $A$ act on the Sylow subgroups of $PA$. If $Q$ is a Sylow subgroup of $PA$, then it is conjugate to $P$. This there is some $pa\in PA$ such that
$$ Q = P^{pa}=P^a $$
Thus $A$ acts transitively on the Sylow subgroups of $PA$, and by the orbit-stabilizer theorem again, $n_p(PA)=[A:N_A(P)]$. 
Edit: below is my original argument for the second part. 
For your second claim, note that $N_G(PA)=N_G(P)A$. This follows, for example, from Frattini's argument applies to $PA\lhd N_G(PA)$. But then we have
\begin{align*}
|N_G(PA)| &= |N_G(P)A|\\
&= \frac{|N_G(P)|\cdot|A|}{|N_G(P)\cap A|}
\end{align*}
Note that the subgroup in the denominator there is just $N_A(P)$. So we then have from above
\begin{align*}
n_p(PA) &= [N_G(PA):N_G(P)]\\
&= \frac{|N_G(PA)|}{|N_G(P)|}\\
&= \frac{|N_G(P)|\cdot|A|}{|N_G(P)|\cdot|N_A(P)|}\\
&= \frac{|A|}{|N_A(P)|}\\
&= [A:N_A(P)]
\end{align*}
A: Additional remark and further refinement, a Theorem of Marshall Hall:
Theorem (M. Hall, 1967) Let $A \unlhd G$, $P \in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where 
$a_p = \#Syl_p(G/A)$
$b_p=\#Syl_p(A)$ and 
$c_p=\#Syl_p(N_{PA}(P \cap A)/(P \cap A))$.
