jmc

We take the inequalities one at a time. Adding $144$ to both sides of the first inequality, we get $$144\ge 54p,$$implying $$\frac{144}{54}\ge p.$$Reducing the fraction and switching the sides (along with the direction of the inequality), we get $p\le \frac{8}{3}$.

To solve the second inequality, we add $20p$ to both sides: $$20p>12$$Dividing both sides by $20$, we get $$p>\frac{12}{20}.$$Reducing the fraction gives $p>\frac{3}{5}$.

We are looking for $p$ which satisfy both inequalities. The intersection of the solutions above is $\boxed{\left(\frac{3}{5},\frac{8}{3}\right]}$.