(3 The base $ABCD$ of the square pyramid $P-ABCD$ is a rhombus with a top angle of $60^{\circ}$, and the dihedral angles between the lateral faces and the base are all $60^{\circ}$. There is a point inside the pyramid that is equidistant from the base and each lateral face, with a distance of 1. Then the volume of the pyramid is ( ).
(A) $16 \sqrt{3}$
(B) $12 \sqrt{3}$
(C) $8 \sqrt{3}$
(D) $4 \sqrt{3}$
(3) It is easy to know that plane $P B D$ is perpendicular to plane $A B C D$, and divides the quadrilateral pyramid into two equal-volume tetrahedra. If the distance from point $G$ to the base and each side face is 1, then $G$ lies on the perpendicular bisector $P H$ of the isosceles triangle $P B D$. Let the projection of $G$ on plane $P A D$ be $E$, and plane $G E H$ passes through vertex $P$ and intersects $A D$ at point $F$. Then $G, E, F, H$ are concyclic, $\angle E F H = 60^{\circ}, \angle E G H = 120^{\circ}$, thus
$$
E H = E F = F H = \sqrt{3},
$$

From this, the height of the equilateral triangle $A B D$ is $A H = 2 \sqrt{3}$, so $A B = 4, S_{\triangle A B D} = 4 \sqrt{3}$,
$$
\begin{array}{l} 
P H = H F \tan 60^{\circ} = 3, \text{ then } \\
V_{P-A B D} = \frac{1}{3} P H \cdot S_{\triangle A B D} = 4 \sqrt{3},
\end{array}
$$

Therefore, $V_{P-A B C D} = 8 \sqrt{3}$. Hence, the answer is C.