The base case, $n = 0$: $$23^0+31^0+46=48$$ and $48 \bmod 48 = 0$.
Inductive Hypothesis: Let's assume it is true for $n = k$. Then $$p(k) = 23^{2k} + 31^{2k}+46$$ $$ \Rightarrow p(k+1) = 23^{2\left(k+1\right)} + 31^{2\left(k+1\right)}+46 = 529\left(23^{2k}\right)+961\left(31^{2k}\right)+46$$
However, I am not sure how to proceed from here in order to get an expression that is divisible by $48$. Thanks in advance for any hints.