 Solution

Let $X$ count the four-vertex sets spanning a clique. Its expectation is $\mu =\left(\genfrac{}{}{0ex}{}{n}{4}\right)p^6$. If $p\ll n^{-2/3}$, then $\mu \to 0$ and Markov's inequality gives $P(X>0)\to 0$.

Now suppose $p\gg n^{-2/3}$. Then $\mu \to \infty$. Two distinct clique indicators are independent unless their vertex sets share at least two vertices. An overlap of two vertices gives $O(n^6)$ ordered pairs requiring eleven distinct edges; an overlap of three gives $O(n^5)$ pairs requiring nine edges. ConsequentlyThe preceding second-moment bound therefore gives $P(X=0)\to 0$. Thusis a threshold for the appearance of a four-clique. Sharing just one vertex creates no dependence because the two edge sets are disjoint.