答案：D
連結 $AC$ 及 $BD$。
$\begin{array}{rcll}
\angle BDC & = & \angle CBQ & \text{(內錯弓形的圓周角)} \\
\angle BDC & = & 39^\circ \\
\angle ADB & = & \angle ADC -\angle BDC \\
\angle ADB & = & 79^\circ -39^\circ\\
\angle ADB & = & 40^\circ
\end{array}$
另外，
$\begin{array}{rcll}
\angle ACB & = & \angle ADB & \text{(同弓形內的圓周角)} \\
\angle ACB & = & 40^\circ
\end{array}$
加上，
$\begin{array}{rcll}
\angle ACD & = & \angle DAP & \text{(內錯弓形的圓周角)} \\
\angle ACD & = & 42^\circ
\end{array}$
所以，可得
$\begin{array}{rcl}
\angle BCD & = & \angle ACD +\angle ACB \\
\angle BCD & = & 42^\circ +40^\circ \\
\angle BCD & = & 82^\circ
\end{array}$
