[Solution] Draw the diameter $A T$ of the circle, intersecting $P Q$ and $B C$ at points $O$ and $H$, respectively. Extend $A B$ and $A C$ to meet the tangent line through $T$ at points $E$ and $F$. Then, the equilateral $\triangle A E F$ is externally tangent to the smaller circle at points $P, T, Q$.

Therefore, $\triangle P Q T$ is also an equilateral triangle. And $\triangle A P Q$ is also an equilateral triangle.

Since they share the common side $P Q$, $\triangle A P Q \cong \triangle P Q T$, thus $A O = O T$, so $O$ is the center of the larger circle.
$A H$ is the altitude from $A$ to side $B C$ in $\triangle A B C$, so
$$
A O = \frac{2}{3} A H.
$$

Also, since $\triangle A P Q \sim \triangle A B C$, we have $\frac{A O}{A H} = \frac{P Q}{B C}$, which means $P Q = \frac{2}{3} B C = 8$.
Therefore, the answer is $(C)$.
$20 \cdot 60$ As shown in the figure, an equilateral triangle $A B C$ is inscribed in a circle. Another smaller circle is internally tangent to the larger circle at $T$, and is tangent to sides $A B$ and $A C$ at points $P$ and $Q$, respectively. If the length of side $B C$ is 12, then the length of segment $P Q$ is
(A) 6.
(B) $6 \sqrt{3}$.
(C) 8.
(D) $8 \sqrt{3}$.
(E) 9.