jmc

Let $F$ be the center of the square base. Since the pyramid is a right pyramid, triangle $PFB$ is a right triangle. The perimeter of the base of the pyramid is 24 inches, so the length of each side of the base is $6$ inches. Since $F$ is the center of the base, $FB$ is half the diagonal of the base, or $(6\sqrt{2})/2=3\sqrt{2}$ inches. Applying the Pythagorean Theorem to triangle $PFB$ gives $$PF=\sqrt{P{B}^2-F{B}^2}=\sqrt{81-18}=\sqrt{63}=\boxed{3\sqrt{7}}\text{ inches}.$$