smc

Firstly note that $p\le \frac{3}{4}$ and $p\le \frac{2}{3}$, as clearly the probability that both $A$ and $B$ occur cannot be more than the probability that $A$ or $B$ alone occurs. The more restrictive condition is $p\le \frac{2}{3}$, since $\frac{2}{3}<\frac{3}{4}$. Furthermore, by the Inclusion-Exclusion Principle, we also have $$\text{P}(A^{\prime }\wedge B^{\prime })=1-\text{P}(A)-\text{P}(B)+\text{P}(A\wedge B)=1-\frac{3}{4}-\frac{2}{3}+p=p-\frac{5}{12},$$ and as a probability must be non-negative, $p-\frac{5}{12}\ge 0$, so $p\ge \frac{5}{12}$. Therefore, combining our inequalities gives $\frac{5}{12}\le p\le \frac{2}{3}$, or $\boxed{[\frac{5}{12},\frac{2}{3}]}$.