Suppose $G$ has order $p^n \cdot m$, where $p$ doesn't divide $m$. Then if $K$ is a Sylow $p$-subgroup of $G$, it has order $p^n$, the highest power of $p$ possible.
Since $K$ is a subgroup of its normalizer $N(K)$, the order of $N(K)$ must be a multiple of $p^n$.
I have no idea if $N(K)$ is normal in $G$, but if it is, the quotient $G/N(K)$ has order $\frac{\lvert G \rvert }{\lvert N(K)\rvert}$, where both numerator and denominator are multiples of $p^n$ (in other words, have the largest $p$-part possible).
So that quotient cannot be a multiple of $p$ (exactly the same number of $p$'s divide the top and bottom), which is exactly what the book is asserting.