Finding the zero-order wave-functions of the perturbed state

If I have an unperturbed Hamiltonian $$\hat{H}_0$$ which is four-fold degenerate corresponding to orthonormal eigenfunctions $$\phi_1$$, $$\phi_2$$, $$\phi_3$$, $$\phi_4$$ and I have some perturbing Hamiltonian $$\hat{H}_p$$, how do I find the wave functions of the perturbed state to zero-order?

Am I right in assuming that because we are dealing with the zero-order, I don't need to consider $$\hat{H}_p$$? If this is the case, are the wave functions of the perturbed state to zero-order simply the degenerate states $$\phi_1$$, $$\phi_2$$, $$\phi_3$$, $$\phi_4$$?

If not, what is a general method of obtaining the wave functions of the perturbed state to zero order for time-independent degenerate perturbation theory?

Lets assume that the degeneracy is completly lifted by the perturbation. Then we have four eigenvectors $$v_1,\dots , v_4$$ with different eigenvalues $$\lambda_1, \dots , \lambda_4$$ (that have all split from the degenerate eigenvalue of $$\hat H_0$$ under consideration) of the total Hamiltonian $$\hat H (\alpha) = \hat H_0 + \alpha \hat H_p$$ (where $$\alpha \in \mathbb{C}$$ is a small parameter). Therefore (under suitable assumptions) we have $$v_i = v_i^0 + \alpha v_i^1 + \alpha^2 v_i^2+ o(\alpha^2)$$ for $$i=1,\dots, 4$$. The coefficients here are unique up to normalisation of the whole vector. The zero order contributions $$v_i^0$$ ($$i=1, \dots ,4 )$$ must be linearly independent, because they span the degenerate subspace. Therefore you can not just take any vectors from the degenerate subspace as the zeroest order eigenvectors, but you must take these specific $$v_i^0$$.
The correct zero order eigenvectors can be obtained as follows under the assumption that the perturbation lifts all degeneracy at the first order: The zeroest order eigenvectors (up to normalization) are given by the eigenvectors of $$P\hat H_p P$$ where $$P$$ is the projection onto the degenerate eigenspace (of $$\hat H_0$$).