Let $G$ be a finite simple undirected graph. Let $G^c$ be the complement graph of $G$ ( https://en.m.wikipedia.org/wiki/Complement_graph) .
Assume that every maximal (w.r.t. set inclusion ) clique of $G^c$ has the same size and denote this common size of a maximal clique by $\omega (G^c)$.
Let $\Omega (G)$ be the intersection number of $G$ (https://en.m.wikipedia.org/wiki/Intersection_number_(graph_theory) .
Then is it true that $\omega(G^c)-1\le \Omega (G)$ ?
Yes, it's true; what's more, the condition on maximal cliques of $G^c$, and the $-1$ in the inequality, are both unnecessary.
If $\omega(G^c) = k$, then there is a set of vertices $\{v_1, v_2, \dots, v_k\}$ forming a clique in $G^c$, and so none of the edges $v_iv_j$ exist in $G$. Therefore in any cover of $G$ by cliques, we are forced to put each vertex $v_i$ in its own clique, and so at least $k$ cliques are needed. This means that $\Omega(G) \ge k$.