6. In the tetrahedron $P-ABC$, $\triangle PAC$ and $\triangle ABC$ are both equilateral triangles with side lengths of 6. If the dihedral angle $P-AC-B$ is $120^{\circ}$, then the surface area of the circumscribed sphere of the tetrahedron $P-ABC$ is $\qquad$
6. $84 \pi$.

As shown in Figure 3, take the midpoint $D$ of $AC$, and connect $DP$, $DB$.
By the problem, we know that $PD \perp AC$, $BD \perp AC$, and the dihedral angle $P-AC-B$ has a plane angle $\angle PDB = 120^{\circ}$.

Let $O$ be the center of the circumscribed sphere of the tetrahedron $P-ABC$, and $O_1$, $O_2$ be the centers of $\triangle ABC$ and $\triangle PAC$, respectively.
Then $OO_1 \perp$ plane $ABC$, $OO_2 \perp$ plane $PAC$, and
$O_2D = O_1D = \frac{1}{3}\left(\frac{\sqrt{3}}{2} \times 6\right) = \sqrt{3}$,
$OO_2 = OO_1$.
It is easy to see that points $O$, $O_2$, $D$, and $O_1$ are coplanar. Connect $OD$.
Then $\angle ODO_1 = 60^{\circ}$, $OO_1 = \sqrt{3} DO_1 = 3$.
Thus, the radius of the circumscribed sphere of the tetrahedron $P-ABC$ is
$R = OB = \sqrt{OO_1^2 + O_1B^2} = \sqrt{21}$.
Therefore, the surface area of the circumscribed sphere of the tetrahedron $P-ABC$ is $4 \pi R^2 = 84 \pi$.