Interesting, but quite easy if I am correct. Call the initial point $I$. Call the center of the ball $O$. Call the marked point $P$. Firstly there exist a plane containing the points $PIO$ (when we have yet to move the ball). Now roll the ball along this plane such that points $P,I,O$ stay on the plane, and $P$ moves to a point that is directly to the right of $O$ ball. Now suppose that the distance rolled is $D_1$. Let the circumference of a circle with center $O$ by $x$. Suppose that $kx< D_1<(k+1)x$ (If it is exactly a multiple of $x$, we are done.) It is easy to see that there exists a plane containing $O$ which when rolled, does not move the point $P$ at all. Now roll the ball along this plane such that point $O$ stays on this plane. As we roll the ball along this plane, the distance between $O$ and $I$ increases, and hence the distance $D_2$ moved by rolling between $O$ and the original center of the circle above $I$ increases. At some point, by Pythagorean theorem, $D_2$ must be $(k+ 1\frac{3}{4})x$ and hence we can roll it back to $I$ with $P$ at the highest point. And this problem is a good problem for final exam, in the sense that writing the solution is not as easy as finding the idea for the solution.

Maybe I'm wrong but did you roll the ball among different planes?? Because The only plane we can use is the plane ball is placed on.

Oh I used those planes I mentioned to be the direction which the ball rolles. Like the axis of earth, just that it is a plane and the ball moves in clockwise or anti-clockwise direction. I edited the solution to attempt to make it clearer.

Actually there's an easier to explain approach, at first we move the ball such a way that P comes to the top of it, obviously this can be done while moving the ball less than the length of it circumference ($x$). Let this length be $d$. Then construct an isosceles triangle with base $d$ and sides $x$ (Hence $d$<$x$ this is possible). So moving the ball among the base will take $P$ to the top, and moving it among any other of the sides of the isosceles triangle won't change the top (because the length is $x$) so finally when it comes back to the first point, $d$ is on top.

Yes, my idea was to keep $P$ as a point that does not move.