I'm not sure to well understand your question, so I add a figure.
The figure is a plane section of your sphere passing from $P$. If I well understand the distance that you want is the length of the arc $PB$.( If this is wrong than my answer is wrong)
You know the radius of the cap, that is the arc $AB=\beta$. In this case the arc $PB$ is simply $\dfrac{\pi}{2}-\alpha-\beta$ , where $\alpha$ is the arc that fix the position of $P$ with respect to the equatorial plane of the sphere(its latitude).
If you know as radius of the cap the distance $CB$ than you can find $\beta=\arcsin (CB)$.
The standard directions at any point $P_1( R , \theta_0, \phi_0)$ on the surface of the sphere are
$ u_1 = \begin{pmatrix} - \sin \phi_0 \\ \cos \phi_0 \\ 0 \end{pmatrix}$
$ u_2 = \begin{pmatrix} - \cos \theta_0 \cos \phi_0 \\ - \cos \theta_0 \sin \phi_0 \\\sin \theta_0 \end{pmatrix} $
The reference direction from which $\alpha$ is measured is $u_2$, clockwise. Therefore, the direction along which $P_2$ is located is
$ u = u_2 \cos \alpha + u_1 \sin \alpha $
Now we have to assume that the center of rotation for motion of the moving point (which is actually rotating) is the center of the sphere. This means that the moving point is moving along a great circle of the sphere. A great circle of the sphere is any circle that lies on the surface of the sphere that has its center the center of the sphere.
To parametrize the arc of motion of $P$, we note that it is spanned by $ u $ (the tangential vector) and $ r $ (the radius vector). These two vector are orthogonal, and $r $ is given by
$ r = \begin{bmatrix} \sin \theta_0 \cos \phi_0 \\ \sin \theta_0 \sin \phi_0 \\ \cos \theta_0 \end{bmatrix} $
Therefore, moving away from $P_1$ by an angle $\phi$, gives point $P$ as follows:
$ P = R \bigg( \cos \phi \ r + \sin \phi \ u \bigg) $
To get $P = P_2$ set the value of angle $\phi = \dfrac{d}{R} $.
Best Answer
The new point is just $$C+\frac{r\cdot CP}{|CP|}$$