This is a question from Chapter 16, A Book of Abstract Algebra by Charles Pinter.
Here are the details from the first two parts of the question (with my own solution) that basically lead to the main problem in the title:
Let $K$ be a p-Sylow subgroup of $G$, and $N=N(K)$ be the normalizer of $K$. Let $a \in N$, and suppose the order of $Ka$ in $N/K$ is a power of $p$. Let $S=<Ka>$ be the cyclic subgroup of $N/K$ generated by $Ka$. Then by the Correspondence Theorem, $S \cong S^{*}/K$ where $S^{*} = \{x \in N:f(x) \in S\}$ and $f$ is a homomorphism from $N$ onto $N/K$. Because $S$ is a p-group, hence $S^{*}/K$ is also a p-group. As $S^{*}$ is a p-subgroup of $G$ (because $K$ and $S^{*}/K$ are p-groups), and by definition $K \subseteq S^{*}$ , hence $K=S^{*}$ and it follows that $Ka=K$ (i.e. every element in the kernel of $N$ gets mapped to the identity element $K$ in $N/K$).
Thus by contradiction, there is no element of $N/K$ that has order a power of $p$ because the order of $K$ in $N/K$ should be 1.
At least that is what I think, however the statement in brackets confused me. I thought even the identity element should not have order a power of $p$. Would anyone please explain why? Also, as a self learner in abstract algebra, I would really appreciate helpful comments on the sketch of proof above. Thanks!