3. $\frac{243 \pi}{16}$.

Draw perpendiculars from points $C$ and $A$ to $PD$, then the perpendiculars must intersect $PD$ at a point $Q$, and
$$\angle AQC = \frac{2 \pi}{3}, \quad AQ = CQ.$$

Since $AC = \sqrt{2} AB = 3 \sqrt{2}$, we have
$$\begin{array}{l} AQ = CQ = \frac{AC}{\sin 120^{\circ}} \cdot \sin \frac{180^{\circ} - 120^{\circ}}{2} = \sqrt{6}, \\ DQ = \sqrt{CD^2 - CQ^2} = \sqrt{3}. \end{array}$$

From $PQ^2 = PC^2 - CQ^2$, we get $PC = \frac{3 \sqrt{3}}{2}$.
Let $O$ be the center of the square $ABCD$. Then the center of the circumscribed sphere of the pyramid $O'$ must lie on the line $PO$, and $PO$ intersects the sphere at another point $P'$.
Also, $CO = \frac{1}{2} AC = \frac{3 \sqrt{2}}{2}$, so
$$PO = \sqrt{PC^2 - CO^2} = \frac{3}{2}.$$

Since $P'O = \frac{CO^2}{PO} = 3$, we have
$$R = \frac{1}{2} (PO + P'O) = \frac{9}{4}.$$

Thus, $V_{\text{known}} = \frac{4}{3} \pi R^3 = \frac{243 \pi}{16}$.