Let us label the $p^2$ rows and the $p^2$ columns of the chessboard by all the pairs $(a, b)$ and $(c, d)$, respectively, where $a, b, c, d \in \{0, 1, \dots, p-1\}$. A field lying in the row $(a, b)$ and in the column $(c, d)$ will be called good if and only if
$$ac \equiv b+d \pmod{p}. \qquad (1)$$
Given a pair $(a, b)$, (1) holds for exactly $p$ pairs $(c, d)$. Thus the number of good fields is $p$ in any row and $p^3$ in total.

It remains to show that there are no four good fields with the prohibited property. Suppose on the contrary that
$$a_i c_j \equiv b_i + d_j \pmod{p} \quad \text{whenever } i, j \in \{1, 2\}, \qquad (2)$$
for some pairs $(a_1, b_1) \neq (a_2, b_2)$ and $(c_1, d_1) \neq (c_2, d_2)$. Subtracting (2) with a fixed $i$ yields
$$a_i(c_2 - c_1) \equiv d_2 - d_1 \pmod{p} \quad \text{for } i \in \{1, 2\}, \qquad (3)$$
which (in the same way) leads to
$$(a_2 - a_1)(c_2 - c_1) \equiv 0 \pmod{p}.$$
Thus $a_1 = a_2$ or $c_1 = c_2$. In view of symmetry, we can assume that $c_1 = c_2$. Then (3) implies that $d_1 = d_2$, hence $(c_1, d_1) = (c_2, d_2)$, a contradiction.