Let $d = \gcd(ab, a^2 + pb^2)$ and suppose $d > 1$. Let $q$ be a prime factor of $d$. Because $a^3 = a(a^2 + pb^2) - pb(ab)$ we see that $q \mid a^3$ and so $q \mid a$ as well. As $q \mid a^2 + pb^2$, this implies that $q \mid pb^2$. Because $\gcd(a, b) = 1$ this can only happen if $q = p$.

So far we have shown that $d > 1$ implies $p \mid a$ and $d = p^\alpha$ for some $\alpha > 0$. In particular, if $p \nmid a$, then $d = 1$. Obviously, if $p \mid a$, then $p \mid d$. We continue to assume $d > 1$ and need to show $d = p$. Write $a = \tilde{p}\tilde{a}$, so that
$$d = \gcd(\tilde{p}\tilde{a}b, p^2\tilde{a}^2 + pb^2) = p \cdot \gcd(\tilde{a}b, p\tilde{a}^2 + b^2).$$
Because $\gcd(\tilde{a}, b) = 1$, the argument at the start of this solution shows that $\gcd(\tilde{a}b, p\tilde{a}^2 + b^2) > 1$ would imply that $p \mid b$, which contradicts $\gcd(a, b) = 1$. Hence $\gcd(\tilde{a}b, p\tilde{a}^2 + b^2) = 1$ and so $d = p$.