Ans: C
Join $OB$.
$\begin{array}{rcl}
\angle BOQ & = & 2\times \angle BPQ \\
& = & 2\times 12^\circ \\
& = & 24^\circ
\end{array}$
In $\Delta OBP$,
$\because OB$ and $OP$ are radii of the semicircle $PBQ$,
$\therefore \angle OBP=\angle OPB=12^\circ$.
Join $OC$.
$\because PQ$ is the tangent to the circle $ABC$ at $O$,
$\therefore \angle COP=\angle OBP = 12^\circ$.
Hence, we have
$\begin{array}{rcl}
\angle BOC & = & 180^\circ – \angle BOQ-\angle COP \\
& = & 180^\circ-24^\circ-12^\circ \\
& = & 144^\circ
\end{array}$
Since $ABOC$ is a cyclic quadrilateral, then we have
$\begin{array}{rcl}
\angle BAC & = & 180^\circ-\angle BOC \\
& = & 180^\circ-144^\circ \\
& = & 36^\circ
\end{array}$
