The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is because

The number of oxidation states shown by an element depends on how readily its valence electrons can participate in bonding. For the inner-transition series this ability is governed by the radial extension (i.e. spatial spread) of the f-orbitals.

• In the lanthanoids, the $$4f$$ subshell is well shielded by the filled $$5s$$ and $$5p$$ orbitals. Because the $$4f$$ orbitals are close to the nucleus (“deeply buried”), their electrons are held rather tightly and can participate in bonding only with difficulty. Hence lanthanoids usually exhibit just the +3 state with only a few showing +2 or +4.

• In the actinoids, the $$5f$$ orbitals are less effectively shielded by the outer $$6s$$ and $$6p$$ electrons. Consequently the $$5f$$ orbitals extend farther from the nucleus; their electrons experience weaker nuclear attraction, become more available for bonding, and can be removed or shared to give several oxidation states ranging typically from +3 up to +7.

Therefore the fundamental reason why actinoids display a greater variety of oxidation states than lanthanoids is that the $$5f$$ orbitals lie farther out (are more diffused) than the $$4f$$ orbitals.

Now examine each option:

Option A - “the $$5f$$ orbitals are more buried than the $$4f$$ orbitals”  ⟶  Incorrect; they are less buried.

Option B - Similar angular parts of the wave functions do not directly influence oxidation state multiplicity; incorrect.

Option C - General reactivity does not decide the count of possible oxidation states; incorrect.

Option D - “the $$5f$$ orbitals extend further from the nucleus than the $$4f$$ orbitals”  ⟶  Correct explanation.

Hence, the answer is:
Option D which is: the $$5f$$ orbitals extend further from the nucleus than the $$4f$$ orbitals