Correct Answer: $95^{\circ}$

Solution :

In the given problem, $\angle \mathrm{AOD}$ and $\angle \mathrm{BOC}$ are angles subtended by the chords $\mathrm{AB}$ and $\mathrm{CD}$ at the centre of the circle.
The angle subtended by a chord at the centre is twice the angle subtended by it at any point on the alternate segment of the circle.
$\mathrm{\angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 100^{\circ} = 50^{\circ}}$
$\mathrm{\angle BDC = \frac{1}{2} \angle BOC = \frac{1}{2} \times 70^{\circ} = 35^{\circ}}$
In $\triangle \mathrm{BPD}$,
$\mathrm{\angle BPD = 180^{\circ} -(50^{\circ} + 35^{\circ}) = 95^{\circ}}$
$\angle\mathrm{ BPD}$ and $\angle \mathrm{APC}$ are vertically opposite angles.
$\mathrm{\angle BPD =\angle APC}$
$\mathrm{\angle APC = 95^{\circ}}$
Hence, the correct answer is $95^{\circ}$.