It seems like you want to create another planet's analog of the Earth's right ascension and declination.
I think that the easiest way to do it is with vector calculus. Spherical trigonometry is mathematically equivalent, but it's a nightmare to keep it straight.
You will need to find axis directions X, Y, Z for the planet. Z is the direction of the planet's north pole, X is the direction of its vernal equinox, and Y forms a right-handed triplet with those two: Y = (Z)x(X).
For that, you will need the directions of both the planet's north pole and its orbit's north pole in Earthly coordinates.
The planet's north pole is simple. You can usually find it as right ascension (α) and declination (δ), the celestial version of Earthly longitude and latitude.
$$ Z(planet) = \{ \cos \delta \cos \alpha, \cos \delta \sin \alpha, \sin \delta \} $$
The orbit's north pole is more complicated. It's often given in the Earth's ecliptic coordinates, and indirectly, from the orbit's inclination and ascending-node longitude.
$$ Z(planet-orbit) = \{ \sin i \sin \Omega, - \sin i \cos \Omega, \cos i \} $$
To get from the Earth's ecliptic coordinates to its equatorial ones, you need the Earth's obliquity ε:
$$ X(ecliptic) = \{1, 0, 0 \} ,\ Y(ecliptic) = \{0, \cos\varepsilon, \sin\varepsilon \}, Z(ecliptic) = \{0, -\sin\varepsilon, \cos\varepsilon \} $$
To get from ecliptic to equatorial coordinates, you must do
$$ Z(PO-equatorial) = Z(PO-ecl,1) X(ecliptic) + Z(PO-ecl,2) Y(ecliptic) + Z(PO-ecl,3) Z(ecliptic) $$
where the 1, 2, and 3, are the three components of Z(PO-ecl) = Z(planet-orbit-ecliptic).
Once you have that, then
$$ X(planet) \sin\varepsilon(planet) = Z(planet-orbit) \times Z(planet) ,\ Y(planet) = Z(planet) \times X(planet) $$
So the position in planet coordinates of an object at position n in Earth equatorial coordinates is {n.X(planet), n.Y(planet), n.Z(planet)} and it can be converted to the planet's analog of right ascension and declination if one so desires.
