, 2019

Obviously, the identical function $f(p) = p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.

First we will show that $f(2) = 2$. Taking $q = 2$ and $p$ any odd prime number, we have
$$f(p)^{f(2)} + 2^p = f(2)^{f(p)} + p^2.$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $f(p) = 2$ for any odd prime number $p$.

Taking any two different odd prime numbers $p, q$ we have
$$2^2 + q^p = 2^2 + p^q \Rightarrow p^q = q^p \Rightarrow p = q,$$
contradiction. Hence, $f(2) = 2$.

So for any odd prime number $p$ we have
$$f(p)^2 + 2^p = 2^{f(p)} + p^2.$$
Copy this relation as
$$2^p - p^2 = 2^{f(p)} - f(p)^2. \qquad (1)$$
Let $T$ be the set of all positive integers greater than 2, i.e. $T = \{3, 4, 5, \dots\}$. The function $g: T \to \mathbb{Z}$, $g(n) = 2^n - n^2$, is strictly increasing, i.e.
$$g(n+1) - g(n) = 2^n - 2n - 1 > 0 \qquad (2)$$
for all $n \in T$. We show this by induction. Indeed, for $n = 3$ it is true, $2^3 - 2 \cdot 3 - 1 > 0$. Assume that $2^k - 2k - 1 > 0$. It follows that for $n = k + 1$ we have
$$2^{k+1} - 2(k+1) - 1 = (2^k - 2k - 1) + (2^k - 2) > 0$$
for any $k \ge 3$. Therefore, (2) is true for all $n \in T$.

As consequence, (1) holds if and only if $f(p) = p$ for all odd prime numbers $p$, as well as for $p = 2$.

Therefore, the only function that satisfies the given relation is $f(p) = p$, for all $p \in \mathbb{P}$.

$\boxed{}$