Prove that $n_p(N)$ divides $n_p(G)$

Theorem: Let $G$ act on the sets $\Omega_1,\Omega_2$ transitively, and assume that there is an onto function $\phi:\Omega_1\to \Omega_2$ such that $\phi(g.x)=g.\phi(x)$ for all $g \in G$ and $x \in \Omega_1$. Then $|\Omega_2|$ divides $|\Omega_1|$.

Proof: Let $\alpha\in \Omega_1$ and $g\in G_{\alpha}$ (stabilizer of $\alpha$). Since $$g\phi(\alpha)=\phi(g\alpha)=\phi(\alpha),$$ we have $G_{\alpha}\leq G_{\phi(\alpha)}$. Since $|\Omega_1|=|G:G_\alpha|$ and $|\Omega_2|=|G:G_{\phi(\alpha)}|$, the result follows from this.

Now, Set $\Omega_1=Syl_p(G)$ , $\Omega_2=Syl_p(N)$ and $\phi(P)=P\cap N$. Now, actions of $G$ on the sets by conjugation are transitive and $\phi$ has desired properties so we are done.

Note: If you do not want mention the theorem, you can directly use the idea in the proof to show the result.

@Bhaskar, I promised you a proof - did not find the time to write it down earlier. Although Mesel already gave you one, mine is of a different flavor and sticks more closely to Sylow theory. You already mentioned that you know that Sylow $p$-subgroups of a normal subgroup $N$ of a group $G$, are of the form $P \cap N$, where $P \in Syl_p(G)$. I will assume that you are familiar with the Frattini Argument and will also use the following easy lemma.

Lemma Let $G$ be a a group, $H \leq G$ and $N \trianglelefteq G$. Then for any $g \in G$ we have $(H \cap N)^g=H^g \cap N$.

Proof. Since $H \cap N \subseteq H$, $(H \cap N)^g \subseteq H^g$, and likewise $(H \cap N)^g \subseteq N^g=N$ (the latter equality follows from the fact that $N$ is normal. Hence $(H \cap N)^g \subseteq H^g \cap N$. Conversely, let $x \in H^g \cap N$, then $x \in N$ and $x=g^{-1}hg$ for some $h \in H$. Hence $h= gxg^{-1} \in N$, since $x \in N$ and $N$ is normal. So, $x=g^{-1}hg \in (H \cap N)^g$ and we are done.

Now let us get to the proof of your post.

Theorem Let $G$ be a group, $N \trianglelefteq G$, and $p$ a prime dividing the order of $G$. Then $$\#Syl_p(N) \mid\#Syl_p(G).$$

Proof. Let $P \in Syl_p(G)$, then $P \cap N \in Syl_p(N)$. Owing to the Frattini Argument we obtain $G=N_G(P\cap N)N$. Observe that $N \cap N_G(P \cap N)=N_N(P \cap N)$ and hence $$[G:N_G(P \cap N)]=[N_G(P \cap N)N:N_G(P \cap N)]=[N:N_N(P \cap N)]=\#Syl_p(N).$$ Finally let $g \in N_G(P)$, then the lemma gives $(P \cap N)^g=P^g \cap N =P \cap N.$ It follows that $N_G(P) \subseteq N_G(P \cap N)$, and thus $\#Syl_p(N)=[G:N_G(P \cap N)] \mid [G:N_G(P)]=\#Syl_p(G)$. $\square$