Let $A$ and $B$ be two distinct points on the circle $\Gamma$, not diametrically opposite. The point $P$, distinct from $A$ and $B$, varies on $\Gamma$. Find the locus of the orthocentre of triangle $ABP$.
1. **Define the orthocenter and its properties:**
   The orthocenter \( H \) of a triangle \( \triangle ABP \) is the point where the altitudes of the triangle intersect. An altitude of a triangle is a perpendicular line segment from a vertex to the line containing the opposite side.

2. **Consider the circle \(\Gamma\) and the points \(A\) and \(B\):**
   Let \(\Gamma\) be a circle with points \(A\) and \(B\) on it. Since \(A\) and \(B\) are not diametrically opposite, the line segment \(AB\) is not a diameter of \(\Gamma\).

3. **Varying point \(P\) on \(\Gamma\):**
   Let \(P\) be a point on \(\Gamma\) distinct from \(A\) and \(B\). As \(P\) varies on \(\Gamma\), we need to find the locus of the orthocenter \(H\) of \(\triangle ABP\).

4. **Symmetry and reflection properties:**
   Consider the line \(PH\) where \(H\) is the orthocenter of \(\triangle ABP\). Let this line intersect the circle \(\Gamma\) again at point \(X\). It is a well-known geometric fact that the orthocenter \(H\) and the point \(X\) are symmetric with respect to the line \(AB\). This means that if you reflect \(X\) across the line \(AB\), you will get the orthocenter \(H\).

5. **Reflection of the circle \(\Gamma\):**
   The reflection of the circle \(\Gamma\) across the line \(AB\) will be another circle, which we can denote as \(\Gamma'\). The points on \(\Gamma'\) are the reflections of the points on \(\Gamma\) across the line \(AB\).

6. **Locus of the orthocenter \(H\):**
   Since \(H\) is the reflection of \(X\) across \(AB\), and \(X\) lies on \(\Gamma\), the locus of \(H\) as \(P\) varies on \(\Gamma\) will be the circle \(\Gamma'\), minus the points \(A\) and \(B\) (since \(P\) cannot be \(A\) or \(B\)).

Conclusion:
The locus of the orthocenter \(H\) of the triangle \(\triangle ABP\) as \(P\) varies on \(\Gamma\) is the circle which is the reflection of \(\Gamma\) across the line \(AB\), minus the points \(A\) and \(B\).

The final answer is the circle which is the reflection of \( \boxed{ \Gamma } \) across the line \(AB\), minus the points \(A\) and \(B\).