jmc

Let's write the entire equation in smaller bases. We have $${(2^2\cdot 3)}^3=\frac{(3^2{)}^2}{3}\cdot 2^{12p}$$Then, using laws of exponents, we have $$2^6\cdot 3^3=\frac{3^4}{3}\cdot 2^{12p},$$which means $$2^6\cdot 3^3=3^3\cdot 2^{12p}.$$Dividing both sides by $3^3$, we have $$2^6=2^{12p}.$$Therefore, we have $12p=6$, which means $p=\boxed{\frac{1}{2}}$.