The following proof is from Dummit and Foote's book, $Abstract$ $Algebra$. The proof is well structured and makes sense, save two minor parts. When they say $G/K $ is isomorphic to a subgroup of $S_p$, why is the justification the $1^{st}$ isomorphism theorem and not Cayley's Theorem? Second, why does $k=1\implies H=K$?
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 $H/K$).
Cayley's theorem is for $G$ acting on itself, not for $G$ acting on the left cosets of a subgroup. In your case $G$ acts transitively on the left cosets of $H$, so this induces an injection $\phi: G/K\to S_p$, injective because if $g_1, g_2$ induce the same permutation, then $g_1g_2^{-1}\in K$. That is to say: in $G/K$ the action is given just by taking any representative $g$ of a given $\overline{g}\in G/K$, but then in $G/K$ the identity is $\overline{e}=K$, so two elements inducing the same permuation being in $K$ means they must be equal in $G/K$ (i.e. the kernel of the map from $G/K$ to $S_p$ is trivial). But then an injective homomorphism is the same thing as saying $G/K$ is isomorphic to a subgroup of the target space, finishing the proof.
Your appeal to Cayley's theorem is not entirely meritless: There is a generalized Cayley theorem, attributed to Herstein, which can be directly applied in the given proof:
Let H be a subgroup of a group $G$. Then there is a canonical homomorphism from $G$ into the group of permutations of the cosets of $H$, and the kernel of this homomorphism is in $H$.
For your first question, observe that each $g\in G$ defines a permutation of $S = \{\gamma H\mid \gamma\in G\}$ via the action $g.(\gamma H) := (g\gamma) H$. Since $\left|G : H\right| = \# S = p$ by assumption, you can choose a bijection $S\to\{1,\dots,p\}$, and this defines a homomorphism $G\to S_p$. This homomorphism is precisely your $\pi_H$. Now the first isomorphism theorem precisely says that $$ G/K = G/\ker\pi_H\cong\operatorname{im}\pi_H\subseteq S_p. $$
Secondly, $k$ was defined to be the index of $K$ in $H$. That is, $k = \#\{hK\mid h\in H\}$. But if $k = 1$, this means that $hK = K$ for all $h\in H$; i.e., $H\subseteq K$. Since $K\subseteq H$ as well, this means $H = K$.
(To see that $K\subseteq H$, recall that $K := \ker\pi_H$, meaning that any $\kappa\in K$ acts as the identity on the set $S$ defined above via the action $\kappa.(\gamma H) = (\kappa\gamma)H$. That is, $$ \kappa.(\gamma H) = (\kappa\gamma)H = \gamma H $$ for all $\kappa\in K$, $\gamma\in G$. In particular, $\kappa.H = H$ for all $\kappa\in K$, so that $\kappa\in H$ for all $\kappa\in K$.)