Let $G$ be a finite group and $P$ a Sylow subgroup. Suppose $n_p \geq [G:P]$, then any $g \in G$ is contained in $\langle P, gPg^{-1}\rangle$
I suspect it would be some form of counting argument. Let $|G| = p^nm$. I have no idea how to calculate $|\langle P, gPg^{-1}\rangle|$ but $P(gPg^{-1})\subset \langle P, gPg^{-1}\rangle $ and $|P(gPg^{-1})| = \frac{p^{2n}}{|P\cap(gPg^{-1})|}$, so need to figure out the size of the intersection. It is clear that $|P\cap(gPg^{-1})|\geq \frac{p^n}{m}$.
Let $m = r + kp, r<p$. Then we have $kp+p+1$ Sylow subgroups. If I can count the number of elements in Sylow subgroups I might be able to make a minimum size of intersection and then force $\langle P, gPg^{-1}\rangle$ to be the whole group. And I am stuck at counting.
Per comment: since $n_p \geq [G:P]$, if the inequality is strict, there exists $g_1, g_2 \notin P$ with $g_1Pg_1^{-1} \neq g_2Pg_2^{-1}$ while $g_1P = g_2P$. Then $g_1Pg_1^{-1} = g_2Pg_1^{-1}$. But since $g_1P = g_2P$ we have $g_1 = g_2p'$ for some $p' \in P$, so $g_1^{-1} = p''g_2^{-1}$, which means $g_1Pg_1^{-1} = g_2Pg_2^{-1}$, a contradiction. Then $n_p = [G:P]$.
Best Answer
Let $P \in Syl_p(G)$. Since $\#Syl_p(G)=n_p(G)=|G:N_G(P)| \geq |G:P|$, and $P \subseteq N_G(P)$, we get $P=N_G(P)$. Fix $g \in G$ and put $H=\langle P, P^g \rangle$. Obviously, $P$ and $P^g$ are two Sylow $p$-subgroups of $H$. Hence by the Sylow theorems (applied in $H$!), there is an $h \in H$, such that $P=P^{gh}$, that is, $gh \in N_G(P)=P \subseteq H$. So $g \in H$ $\square$
Note If for a subgroup $J \leq G$ it holds that for every $g \in G$, $g \in \langle J,J^g \rangle$, the subgroup $J$ is called abnormal. So, above we proved that for a Sylow $p$-subgroup with $P=N_G(P)$, $P$ is abnormal. In fact, the converse is also true:
This last result explains the name "abnormal": if $J$ is a (proper) abnormal subgroup of $G$, then there exists no proper normal subgroup between $J$ and $G$.