Let $ABC$ be a triangle.  Point $P$ lies in the interior of $\triangle ABC$ such that $\angle ABP = 20^\circ$ and $\angle ACP = 15^\circ$.  Compute $\angle BPC - \angle BAC$.
1. Let $\angle BAC = \theta$. We need to find $\angle BPC - \angle BAC$.
2. Since $P$ lies inside $\triangle ABC$, we can use the fact that the sum of angles around point $P$ is $360^\circ$.
3. We know $\angle ABP = 20^\circ$ and $\angle ACP = 15^\circ$.
4. Let $\angle PBC = x$ and $\angle PCB = y$.
5. Using the angle sum property of a triangle, we have:
   \[
   \angle PBC + \angle BPC + \angle PCB = 180^\circ
   \]
6. Also, in $\triangle ABP$ and $\triangle ACP$, we have:
   \[
   \angle PBA = 180^\circ - \angle ABP - \angle BAP = 180^\circ - 20^\circ - \theta = 160^\circ - \theta
   \]
   \[
   \angle PCA = 180^\circ - \angle ACP - \angle CAP = 180^\circ - 15^\circ - \theta = 165^\circ - \theta
   \]
7. Since $\angle PBC + \angle PCB + \angle BPC = 180^\circ$, we can write:
   \[
   \angle BPC = 180^\circ - \angle PBC - \angle PCB
   \]
8. Using the exterior angle theorem in $\triangle PBC$, we have:
   \[
   \angle BPC = \angle PBA + \angle PCA = (160^\circ - \theta) + (165^\circ - \theta) = 325^\circ - 2\theta
   \]
9. Therefore, we have:
   \[
   \angle BPC - \angle BAC = (325^\circ - 2\theta) - \theta = 325^\circ - 3\theta
   \]
10. Since $\angle BAC = \theta$, we need to find the difference:
    \[
    \angle BPC - \angle BAC = 325^\circ - 3\theta - \theta = 325^\circ - 4\theta
    \]
11. Given that $\angle BPC = 180^\circ - (\angle PBC + \angle PCB)$, we can substitute:
    \[
    \angle BPC = 180^\circ - (180^\circ - \theta - 35^\circ) = 180^\circ - 180^\circ + \theta + 35^\circ = \theta + 35^\circ
    \]
12. Therefore, the difference is:
    \[
    \angle BPC - \angle BAC = (\theta + 35^\circ) - \theta = 35^\circ
    \]

The final answer is $\boxed{35^\circ}$.