I'm new at this topic, so I'm not sure about whether is my solution acceptable or not. Could you check it?
By Sylow theorem there exist $|P|=7$ and $|Q|=8$. Sylow $p$-subgroup is normal iff it is unique. Denote by $n_p$ amount of different Sylow $p$-subgroups. Suppose $n_i \neq 1$ for $i=2,7$. Since $|G|=n_p \cdot N_G(P)$ and $n_p\equiv 1 \pmod{p}$, we have that $n_7=8$ and $n_2=7$. So there are $8$ conjugates of $P$ which intersect in identity. Thus there are $(7-1)\cdot 8=48$ elements of order $7$ which can't be elements of Sylow 2-subgroups. There left $8$ elements which are obviously not enough to construct $7$ different Sylow 2-subgroups of order $8$. We've got contradiction.