Suppose that $a b c d = n^{2}$ for some $n \in \mathbb{Z}^{+}$. We can suppose that $a < c < d < b$. From this, we have $b d + a c - a d - b c = (b - a)(d - c) > 0$, thus
$$a d + b c < \frac{1}{2}(a d + b c + b d + a c) = \frac{(a + b)(c + d)}{2} = \frac{p^{2}}{2} .$$
Denote $\gcd(a d, b c) = k \in \mathbb{Z}^{+}$ then $a d = k u^{2}$, $b c = k v^{2}$ with $u, v \in \mathbb{Z}^{+}$, and $u < v$, $\gcd(u, v) = 1$. Since $a, b, c, d$ are coprime to $p$ then $\gcd(k, p) = 1$. We have
$$k\left(v^{2} - u^{2}\right) = b c - a d = (p - a) c - a(p - c) = p(c - a) .$$
So $p \mid v^{2} - u^{2}$ since $\gcd(k, p) = 1$. Hence, $p \mid v - u$ or $p \mid v + u$. In both cases, we always have $u + v \geq p$. From this, we can conclude that
$$a d + b c = k\left(u^{2} + v^{2}\right) > k \frac{(u + v)^{2}}{2} \geq \frac{p^{2}}{2}$$
by AM-GM inequality. But this contradicts the inequality we stated above, then $a b c d$ cannot be a perfect square.