Given BPC = 90^, we have:

The angle between a tangent (PC) and a chord (BP) drawn from the point of contact (P) is equal to the angle in the alternate segment.
Therefore, $\angle BPC = \angle BAP$.
Given $\angle BPC = 90^\circ$, we have:

$$\angle BAP = 90^\circ$$

In $\triangle BPC$, we are given $\angle BPC = 90^\circ$ and $\angle C = 25^\circ$.
The sum of angles in a triangle is $180^\circ$.

$$\angle PBC + \angle BPC + \angle C = 180^\circ$$ $$\angle PBC + 90^\circ + 25^\circ = 180^\circ$$ $$\angle PBC + 115^\circ = 180^\circ$$ $$\angle PBC = 180^\circ - 115^\circ$$ $$\angle PBC = 65^\circ$$

Points A, B, and C are collinear, meaning they lie on the same straight line. Therefore, the angle $\angle ABP$ is the same as $\angle PBC$.

$$\angle ABP = \angle PBC = 65^\circ$$

In $\triangle ABP$, we have:
$\angle BAP = 90^\circ$ (from Step 1)
$\angle ABP = 65^\circ$ (from Step 3)
$\angle APB = y$
The sum of angles in $\triangle ABP$ is $180^\circ$.

$$\angle BAP + \angle ABP + \angle APB = 180^\circ$$ $$90^\circ + 65^\circ + y = 180^\circ$$ $$155^\circ + y = 180^\circ$$ $$y = 180^\circ - 155^\circ$$ $$y = 25^\circ$$

The value of $y$ is \boxed{\text{25^\circ}}.