smc

Notice that $\frac{{\log }_b({\log }_{b}a)}{{\log }_{b}a}$ strongly resembles the change of base rule. Recall that ${\log }_{b}a=\frac{{\log }_{c}a}{{\log }_{c}b}$. Taking the base on the RHS to be $b$, we get that $p={\log }_a({\log }_{b}a)$. Raising $a$ to both sides, we get that $$a^p=a^{{\log }_a({\log }_{b}a)}$$ $$=\boxed{{\log }_{b}a}$$