We are given $a^2 + p^2 = b^2$, where $a$, $b$ are positive integers and $p$ is a prime number.

Rewrite the equation:
$$b^2 - a^2 = p^2$$
$$(b - a)(b + a) = p^2$$

Since $p$ is prime, $p^2$ has only three positive divisors: $1$, $p$, $p^2$.
So, the pairs $(b - a, b + a)$ must be $(1, p^2)$, $(p, p)$, or $(p^2, 1)$.
But $b + a > b - a$, so only $(b - a, b + a) = (p, p)$ or $(1, p^2)$ are possible.

Case 1: $b - a = p$, $b + a = p$

Then $b - a = b + a$ implies $a = 0$, which is not allowed since $a$ is positive.

Case 2: $b - a = 1$, $b + a = p^2$

Then:
$$b - a = 1 \implies b = a + 1$$
$$b + a = p^2$$
Add the two equations:
$$(b - a) + (b + a) = 1 + p^2 = 2b \implies 2b = p^2 + 1 \implies b = \frac{p^2 + 1}{2}$$
Subtract the first from the second:
$$(b + a) - (b - a) = p^2 - 1 = 2a \implies a = \frac{p^2 - 1}{2}$$

Now, $a$ and $b$ are positive integers for any odd prime $p$ (since $p^2$ is odd, so $p^2 + 1$ and $p^2 - 1$ are even).

Now, compute $2(b + p)$:
$$2(b + p) = 2 \left( \frac{p^2 + 1}{2} + p \right ) = (p^2 + 1) + 2p = p^2 + 2p + 1 = (p + 1)^2$$

Thus, $2(b + p)$ is a perfect square, as required.