1. Define Points and Properties:
Let $C'$ and $B'$ be points on $AB$ and $AC$, respectively, such that $BB' = BC$ and $CC' = CB$. Let $X$ and $Y$ be the midpoints of $BB'$ and $CC'$, respectively.

2. Concyclic Points:
Since $A, P, Q,$ and $H$ are concyclic, the points $P$ and $Q$ lie on the circumcircle of $\triangle APQH$. This implies that $\angle APH = \angle AQH$.

3. Spiral Similarity:
The triangle $\triangle HPQ$ has a fixed shape due to the concyclic condition. By the properties of spiral similarity, the midpoint $K$ of $PQ$ lies on a fixed line.

4. Special Cases:
- For $P = B$, the point $K$ coincides with $X$, the midpoint of $BB'$.
- For $Q = C$, the point $K$ coincides with $Y$, the midpoint of $CC'$.

5. **Line Segment $XY$:**
Since $K$ lies on the line segment $XY$, we need to analyze the positions of $X$ and $Y$ relative to the circle $\odot(BC)$.

6. Angles and Midpoints:
Note that $\angle BMB' = \angle BXC$ and $\angle CMC' = \angle BYC$, where $M$ is the midpoint of $BC$.

7. Angle Conditions:
Given that $\min \{\angle B, \angle C\} \ge 60^\circ$, we have $BB' > B'C$ and $CC' > C'B$. This implies that points $X$ and $Y$ lie entirely within the circle $\odot(BC)$.

8. Conclusion:
Since $K$ lies inside the circle $\odot(BC)$, the angle $\angle BKC$ must be greater than $90^\circ$.

$$\boxed{\angle BKC > 90^\circ}$$