$ABCD$ is a cyclic quadrilateral and the sides $DC$ and $AB$ are produced to meet at $P$. If $\angle APD=25^{\circ }$ and $\angle PBC=70^{\circ }$, then $\angle BAD$ is equal to:

In a cyclic quadrilateral, the exterior angle at any vertex is equal to the interior opposite angle.
Here, side $AB$ is extended to $P$, making $\angle PBC$ an exterior angle at vertex $B$.
The interior opposite angle to $\angle PBC$ is $\angle ADC$.

$$\angle ADC=\angle PBC=70^{\circ }$$

Now, consider $△APD$. The sum of angles in a triangle is $180^{\circ }$:

$$\angle BAD+\angle ADC+\angle APD=180^{\circ }$$

$$\angle BAD+70^{\circ }+25^{\circ }=180^{\circ }$$

$$\angle BAD+95^{\circ }=180^{\circ }$$

$$\angle BAD=180^{\circ }-95^{\circ }=85^{\circ }$$