Show that for every $g \in G$, $g$ lies in the subgroup generated by $P$ and $g^{−1}Pg$. 
Let $G$ be a finite group and let $P$ be a Sylow $p$-subgroup of $G$. Suppose that the number of Sylow $p$-subgroups of $G$ is at least $|G : P|$, where $|G : P|$ is the index of $P$ in $G$. Show that for every $g \in G$, $g$ lies in the subgroup generated by $P$ and $g^{−1}Pg$.


My attempt:
Let $n_p$ denote the number of Sylow $p$-subgroups contained in $G$, then we know $n_p=\frac{|G|}{p^n}$ where $|G|=p^n\times q$ with $\gcd(p,q)=1$.
I have tried to let $G$ act on the left cosets $\{P,a_1P,...,a_{q-1}P\}$ and I get nothing else useful except there is a bijection between stabilizers of $a_iP$ and Sylow $p$-subgroups of $G$.
How am I supposed to prove that $g\in\langle P,g^{-1}Pg\rangle$?

Fun thing to observe: a general converse is also true. For Sylow $p$-groups this yields an if and only if so to speak.
Proposition Let $J$ be a subgroup of the group $G$ with the property that for every $g \in G$, $g \in \langle J,J^g \rangle$. Assume that $J \leq H \leq G$, then $N_G(H)=H$.

Proof Let $g \in N_G(H)$. Since $J \leq H$, we also have $J^g \leq H^g=H$. Hence $\langle J, J^g\rangle \leq H$. And the property of $J$ guarantees $g \in H$. 
Note The group $J$ with the property above is sometimes called abnormal. Observe that if $P \in Syl_p(G)$, then $N_G(P)$ is abnormal.
