The more standard notation for $K_{p,n}$ is $O^{p'}(A_n)$. It is the smallest normal subgroup whose index is not divisible by $p$. For each one you need to decide if (1) the number is the order of a normal subgroup, (2) the index is not divisible by $p$, and (3) if it is the smallest number like that. For $n=4$, this is a very small group, so just list all elements of $U_{p,n}$. For $n \geq 5$, $A_n$ has very few normal subgroups so the answer is fairly simple.
The complete answer is:
$$K_{p,n} = \begin{cases}
1 & \text{ if } \max(p,3) > n \text{ or } (p,n) = (2,3)\\
A_n & \text{ if } \max(p,5) \leq n \text{ or } (p,n) = (3,3) \text{ or } (p,n) = (3,4) \\
K_4 & \text{ if } (p,n) = (2,4)
\end{cases}$$
where $K_4$ is the Klein four-group.
This says: If $n$ is too small, then $K_{p,n}=1$ since $A_n$ has only the identity Sylow $p$-subgroup. If $n$ is big enough, $K_{p,n}=A_n$, since it is a non-identity normal subgroup. On the border you have to check cases, and $(p,n)=(2,4)$ is actually different.