Solution:

Let $\Gamma'$ be the circle symmetric to $\Gamma$ with respect to $A B$, and let $A'$ and $B'$ be the 2 points of $\Gamma'$ such that $A A'$ and $B B'$ are perpendicular to $A B$. We will show that the required locus consists of the circle $\Gamma'$ minus the points $A'$ and $B'$.
Let us consider the longer of the 2 arcs of $\Gamma$ bounded by $A$ and $B$, and let $\alpha$ be the angle under which the points of that arc see the segment $A B$. With reference to the figure above, in order to show that $H$ belongs to $\Gamma'$ it suffices to show that $H$ lies above the line $A B$ and $A \widehat{H} B=180^{\circ}-\alpha$, or that $H$ lies below the line $A B$ and $A \widehat{H} B=\alpha$. Let us now denote by $R$ and $S$ the feet of the altitudes drawn from $A$ and $B$, respectively, and let us distinguish several cases.

- Case 1: $P$ lies above $A B$ and $A P B$ is acute-angled (figure on the left). In this case $H$ lies above the line $A B$, and by examining the quadrilateral $P S H R$, which has 2 right angles, we get that $A \widehat{H} B=S \widehat{H} R=180^{\circ}-\alpha$.

- Case 2: $P$ lies above $A B$ and $A P B$ is right-angled. In this case $H$ trivially coincides with $A$ or $B$.

- Case 3: $P$ lies above $A B$ and $A P B$ is obtuse-angled at $A$ (figure in the center). In this case $H$ lies below $A B$. Moreover the triangles $A R P$ and $A S H$ are right-angled and have the angles at $A$ vertically opposite. It follows that $A \widehat{H} B=A \widehat{P} B=\alpha$. An entirely analogous argument holds if $A P B$ is obtuse-angled at $B$.

- Case 4: $P$ lies below $A B$ (figure on the right). In this case $A P B$ is obtuse-angled at $P$ and $A \widehat{P} B=180^{\circ}-\alpha$. In this case too $H$ lies below $A B$, and by examining the quadrilateral $P R H S$, which has 2 right angles, one obtains that $A \widehat{H} B=\alpha$.

Conversely, let $H$ be any point of $\Gamma'$. Let us define $P$ as the orthocenter of $A B H$. By an argument entirely analogous to the previous one (in which one simply exchanges the roles of $\Gamma$ and $\Gamma'$, as well as of $P$ and $H$) we get that $P$ belongs to $\Gamma$, and is different from $A$ and $B$ provided that $H$ is different from $A'$ and $B'$. Since $P$ is the orthocenter of $A B H$, by a well-known property we have in turn that $H$ is the orthocenter of $A B P$. This shows that every point of $\Gamma'$ minus the points $A'$ and $B'$ belongs to the required locus.

Solution:

Let us denote by $C$ the point on $\Gamma$ diametrically opposite to $B$. Since $B C$ is a diameter, the angle $C \widehat{A} B$ is right: it follows that the lines $P H$ and $C A$ are parallel, since both are perpendicular to $A B$. For the same reason, $C \widehat{P} B$ is also right, hence $C P$ and $A H$ are also parallel, since both are perpendicular to $P B$.
It follows that $A H P C$ is a parallelogram, and therefore $P H=A C$ for every $P$ on $\Gamma$. Moreover $P H$ always has direction perpendicular to $A B$, and hence the vector $\overrightarrow{P H}$ is constant.
From this we deduce that the required locus is given by the circle $\Gamma'$, which is the translate of $\Gamma$ by the vector $\overrightarrow{P H} \equiv \overrightarrow{C A}$, minus the translates of $A$ and $B$, which are precisely $A'$ and $B'$.

Solution:

Let us denote by $O$ and $r$ the center and the radius of the circle $\Gamma$. Let us first determine the set formed by the centroids $G_{P}$ of the triangles $A B P$ as $P$ moves along $\Gamma$. Let $M$ be the midpoint of $A B$. Then $G_{P}$ is given by the image of $P$ under the homothety of center $M$ and ratio $\frac{1}{3}$. Therefore the set of points $G_{P}$ forms the circle $\Gamma'$ of center $O'$ and radius $\frac{r}{3}$, where $O'$ is the point of the segment $O M$ such that $M O'=\frac{1}{3} M O$ (and deprived of the two points corresponding to $A$ and $B$ under the homothety). At this point, to obtain the desired set we need to apply to $\Gamma'$ the homothety of center $O$ and ratio 3. Indeed, in every triangle the centroid $G$ divides the segment having as endpoints the orthocenter $H$ and the circumcenter $C$ into two parts such that $G H=2 G C$, and in our case the triangles $A B P$ all have the same circumcenter $O$, which is fixed. Therefore one obtains the circle $\Gamma''$ (deprived of two points) having radius $r$ and center at the point $O''$, where $O''$ is the point on the ray with origin $O$ passing through $M$, placed at distance $2 O M$ from $O$. In other words, $\Gamma''$ is the circle symmetric to $\Gamma$ with respect to $M$, deprived of two points.