Point $A$ shall be projected on a plane with the height $k$ which is parallel to the xy axis. The new point needs to have the same distance to the origin $ r $ and azimuth $ \theta $ on the xy plane. Searched for is the correct $\phi $.
I used the following formulas to convert the cartesian coordinates to it's spherical representation.
$$ Carthesian: (x,y,z) = A \\ Spherical: (r,\theta,\phi) $$
Spherical to Carthesian
$$ x = r * \sin(\phi) * \cos(\theta)\\ y = r * \sin(\phi) * \sin(\theta) \\ z = r * \cos(\phi) \\ $$
Carthesian to Spherical
$$ r = \left\lvert \left\lvert A \right\rvert \right\rvert = \sqrt{ x^{ 2 } + y^{ 2 } + z^{ 2 } } \\ \theta = \arctan{\frac{ y }{ x }} \\ \phi = \arccos(\frac{ z }{ x }) \\ r = \textrm{radial distance}\\ \theta = \textrm{inclination (elevation)}\\ \phi = aziumuth $$
I tried to solve several equations with the assumption that the magnitude and $ \theta $ of both vectors need to be the same, the y of point B is known (k) as well well as xyz of point A, but did not manage to solve these. My next best bet would be to create a linear equation and trying to find the intersection with the plane, but I haven't worked with spherical coordinates before and am a bit lost here.
The easiest way to determine $\phi_B$ seems to use the $Z$ coordinates. For the plane $z=k$, while for point B $z=r\cos(\phi_B)$, so: $$k=r\cos(\phi_B) \rightarrow \phi_B=\arccos(k/r)$$ (where $r=r_A=r_B$ is known). While also $\theta=\theta_A=\theta_B$ is given.
So if you need the coordinates in "XYZ", u can simply use the conversion rules you wrote up.