jmc

Let $F$ be the center of the square base. Since the pyramid is a right pyramid, segment $\overline{PF}$ is an altitude of triangle $PBD$. Since $PBD$ is an equilateral triangle with side length 6, $PFB$ is a 30-60-90 triangle with $FB=BD/2=3$ and $PF=3\sqrt{3}$. Finally, $\overline{BD}$ is a diagonal of square base $ABCD$, so we have $BC=BD/\sqrt{2}=6/\sqrt{2}=3\sqrt{2}$. Therefore, the volume of the pyramid is $$\frac{[ABCD](PF)}{3}=\frac{(3\sqrt{2}{)}^2(3\sqrt{3})}{3}=\boxed{18\sqrt{3}}.$$