 Solution

The unperturbed $n=2$ level is degenerate. The useful point is that a central perturbation is already diagonal within its usual orbital eigenbasis:Angular orthogonality removes the possible mixing of $2s$ and $2p$, and also the mixing of different magnetic sublevels. Therefore the matrix required by degenerate perturbation theory needs no further diagonalization: its diagonal entries are precisely the expectation values above. It would be incorrect to justify the calculation by claiming that the unperturbed spectrum was nondegenerate.

For $b=0$ the energy needs only $n$. For a nonzero central confinement term the energy generally needs $n,l$, but remains independent of $m$. The magnetic quantum number is still needed to specify an individual orbital eigenstate, rather than just its energy.