Solution:

The given equality may be written as
$$(a-b)\left(a^{2}+a b+b^{2}\right)=4 p^{2}$$
Since $a-b<a^{2}+a b+b^{2}$, it follows from (1) that
(2) $a-b<2 p$.
Now consider two cases:

1. $p=2$, and
2. $p$ is an odd prime.

Case 1: $p=2$. Then (1) becomes
(3) $(a-b)\left(a^{2}+a b+b^{2}\right)=16$ :
In view of (2) and (3), it must be $a-b=1$ or $a-b=2$. If $a-b=1$, then substituting $a=b+1$
in (3) we obtain
$$b(b+1)=5$$
which is impossible since $b(b+1)$ is an even integer.
If $a-b=2$, then substituting $a=b+2$ in (3) we get
$$3 b(b+2)=4$$
which is obviously impossible.

Case 2: $p$ is an odd prime. Then (1) yields $a-b \mid 4 p^{2}$. This together with the facts that
$p$ is a prime and that by (2) $a-b<2 p$, yields $a-b \in\{1,2,4, p\}$.

If $a-b=1$, then substituting $a=b+1$ in (1) we obtain
$$3 b(b+1)+1=4 p^{2}$$
which is impossible since $3 b(b+1)+1$ is an odd integer.

If $a-b=2$, then substituting $a=b+2$ in (1) we obtain
$$3 b^{2}+6 b+4=2 p^{2}$$
whence it follows that
$$2\left(p^{2}-2\right)=3\left(b^{2}+2 b\right) \equiv 0(\bmod 3)$$
and hence
$$p^{2} \equiv 2(\bmod 3)$$
Since $p^{2} \equiv 1(\bmod 3)$ for each odd prime $p>3$ and $3^{2} \equiv 0(\bmod 3)$, it follows that the congruence (5) is not satisfied for any odd prime $p$.

If $a-b=4$, then substituting $a=b+4$ in (1) we obtain
$$3 b^{2}+12 b+16=p^{2}$$
whence it follows that $b$ is an odd integer such that
$$3 b^{2} \equiv p^{2}(\bmod 4)$$
whence since $p^{2} \equiv 1(\bmod 4)$ for each odd prime $p$, we have
$$3 b^{2} \equiv 1(\bmod 4)$$
However, since $b^{2} \equiv 1(\bmod 4)$ for each odd integer $b$, it follows that the congruence (6) is not satisfied for any odd integer $b$.

If $a-b=p$, then substituting $a=b+p$ in (1) we obtain
$$p\left(3 b^{2}+3 b p+p^{2}-4 p\right)=0$$
i.e.,
(7)
$$3 b^{2}+3 b p+p^{2}-4 p=0$$
If $p \geq 5$, then $p^{2}-4 p>0$, and thus (7) cannot be satisfied for any positive integer $b$. If $p=3$, then $(7)$ becomes
$$3\left(b^{2}+3 b-1\right)=0$$
which is obviously not satisfied for any positive integer $b$.

Hence, there does not exist positive integers $a$, $b$ and a prime $p$ such that $a^{3}-b^{3}=4 p^{2}$.