Understanding an elementary proof from Dummit and Foote
When they say $G/K$ is isomorphic to a subgroup of $S_p$, why is the justification the 1st isomorphism theorem and not Cayley's Theorem?
Well, Cayley's theorem uses regular representation of $G$ to show that $G$ embeds in some $S_n$. You don't have that here, you have group action on a set (note that $G/H$ is not a group until we prove $H$ is normal), i.e. $\pi_H\colon G\to \operatorname{Perm}(G/H) = S_p$ is a group homomorphism and by the first isomorphism theorem, $\pi_H'\colon G/K\to S_p$ is monomorphism. Perhaps this can be thought of as some application of Cayley, but there is no reason to. If you tried applying Cayley directly on $G/K$, what you would get is that $G/K$ is subgroup of $S_{kp}$, not $S_p$, which is a crucial difference.
Second, why does $k=1\implies H=K?$
$k$ is defined as $k = |H:K|$. If $k = 1$, then $H/K$ is trivial group which occurs precisely when $H = K$ (if there is $h\in H$, $h\not\in K$, then $hK$ is non-trivial element of $H/K$).