Answer: B
\(B\) and \(C\) are joined.
In \(\triangle BCP\), \(\angle PBC + \angle PCB =\) the exterior angle \(\angle APC\).
or, \(\angle ABC + \angle DCB = 40^\circ\).
or, \(\cfrac{1}{2}\angle AOC + \cfrac{1}{2}\angle BOD = 40^\circ\).
or, \(\angle AOC + \angle BOD = 80^\circ\).
\(B\) and \(C\) are joined.
In \(\triangle BCP\), \(\angle PBC + \angle PCB =\) the exterior angle \(\angle APC\).
or, \(\angle ABC + \angle DCB = 40^\circ\).
or, \(\cfrac{1}{2}\angle AOC + \cfrac{1}{2}\angle BOD = 40^\circ\).
or, \(\angle AOC + \angle BOD = 80^\circ\).