Problem 1398

1. Identify the coordinates of points:
- $O = (0,0)$
- $A = (0,a)$
- $B = (0,b)$
- $\Gamma$ is a circle with diameter $\overline{AB}$, so the center of $\Gamma$ is the midpoint of $\overline{AB}$, which is $\left(0, \frac{a+b}{2}\right)$.

2. **Properties of the circle $\Gamma$:**
- Since $\overline{AB}$ is the diameter, any point $P$ on $\Gamma$ will form a right angle with $\overline{AB}$ at $P$. This is due to the inscribed angle theorem, which states that an angle inscribed in a semicircle is a right angle.

3. **Angle $\angle BPA$:**
- Since $\overline{AB}$ is the diameter, $\angle BPA = 90^\circ$.

4. **Coordinates of point $Q$:**
- Line $PA$ meets the x-axis again at $Q$. Since $A$ is on the y-axis, the line $PA$ will intersect the x-axis at some point $Q$ with coordinates $(x, 0)$.

5. **Angle $\angle BQP$:**
- We need to show that $\angle BQP = \angle BOP$.

6. Right angles in the configuration:
- Since $Q$ and $O$ are on the x-axis, and $B$ is on the y-axis, $\angle BOQ = 90^\circ$.

7. **Cyclic quadrilateral $BPOQ$:**
- To prove that $\angle BQP = \angle BOP$, we need to show that $BPOQ$ is a cyclic quadrilateral.
- Since $\angle BPA = 90^\circ$ and $\angle BOQ = 90^\circ$, we have two right angles.
- In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Here, $\angle BPA + \angle BOQ = 90^\circ + 90^\circ = 180^\circ$.

8. Conclusion:
- Since $\angle BPA$ and $\angle BOQ$ are supplementary, $BPOQ$ is a cyclic quadrilateral.
- Therefore, $\angle BQP = \angle BOP$.

$\blacksquare$