Which of the following statements in relation to the hydrogen atom is correct?

For a one-electron species like the hydrogen atom, the energy of an orbital depends only on the principal quantum number $$n$$ and is independent of the azimuthal quantum number $$\ell$$ (or the subshell label s, p, d, …).
Mathematically, the Bohr-Sommerfeld (or Schrödinger) result gives

$$E_n = -\dfrac{R_H}{n^{2}}$$

where $$R_H$$ is the Rydberg constant for hydrogen. Because $$E_n$$ contains no term involving $$\ell$$, every orbital that has the same $$n$$ must possess exactly the same energy.

For $$n = 3$$ the orbitals are 3s ($$\ell = 0$$), 3p ($$\ell = 1$$) and 3d ($$\ell = 2$$). Since all three share the same principal quantum number, their energies are equal:

$$E_{3s} = E_{3p} = E_{3d}$$

Therefore, none of the subshells 3s, 3p, or 3d is lower or higher in energy than the others in a hydrogen atom.

Hence, the only correct statement among the options is:

Option D which is: 3s, 3p and 3d orbitals all have the same energy