The pair(s) of diamagnetic ions is(are)

For any ion, magnetic behaviour depends on the presence or absence of unpaired electrons.
No unpaired electrons ⇒ diamagnetic, at least one unpaired electron ⇒ paramagnetic.

Lanthanide ground-state configurations are based on the xenon core $$[Xe]$$ followed by $$4f$$, $$5d$$ and $$6s$$ subshells. While writing ionic configurations, electrons are removed first from $$6s$$, then $$5d$$, then $$4f$$.

Case A (La$$^{3+}$$, Ce$$^{4+}$$)

Lanthanum (Z = 57): neutral atom $$[Xe]\,5d^{1}6s^{2}$$.
La$$^{3+}$$ : remove three electrons (two from $$6s$$, one from $$5d$$) ⇒ $$[Xe]\,4f^{0}$$.
All subshells are completely filled; no unpaired electrons ⇒ diamagnetic.

Cerium (Z = 58): neutral atom $$[Xe]\,4f^{1}5d^{1}6s^{2}$$ (the commonly quoted ground state).
Ce$$^{4+}$$ : remove four electrons (two $$6s$$, one $$5d$$, one $$4f$$) ⇒ $$[Xe]\,4f^{0}$$.
Again, no unpaired electrons ⇒ diamagnetic.

Case B (Yb$$^{2+}$$, Lu$$^{3+}$$)

Ytterbium (Z = 70): neutral atom $$[Xe]\,4f^{14}6s^{2}$$ (filled $$4f$$ shell).
Yb$$^{2+}$$ : remove the two $$6s$$ electrons ⇒ $$[Xe]\,4f^{14}$$.
The $$4f$$ subshell is completely filled; no unpaired electrons ⇒ diamagnetic.

Lutetium (Z = 71): neutral atom $$[Xe]\,4f^{14}5d^{1}6s^{2}$$.
Lu$$^{3+}$$ : remove two $$6s$$ and one $$5d$$ electron ⇒ $$[Xe]\,4f^{14}$$.
Filled $$4f^{14}$$ subshell; no unpaired electrons ⇒ diamagnetic.

Case C (La$$^{2+}$$, Ce$$^{3+}$$)

La$$^{2+}$$ : $$[Xe]\,5d^{1}$$ - one electron in $$5d$$ ⇒ unpaired ⇒ paramagnetic.
Ce$$^{3+}$$ : $$[Xe]\,4f^{1}$$ - one electron in $$4f$$ ⇒ unpaired ⇒ paramagnetic.
Hence the pair is not diamagnetic.

Case D (Yb$$^{3+}$$, Lu$$^{2+}$$)

Yb$$^{3+}$$ : $$[Xe]\,4f^{13}$$ - one vacancy (equivalently one unpaired electron) in $$4f$$ ⇒ paramagnetic.
Lu$$^{2+}$$ : $$[Xe]\,4f^{14}5d^{1}$$ - one electron in $$5d$$ ⇒ unpaired ⇒ paramagnetic.
Hence this pair is not diamagnetic.

Therefore, the diamagnetic pairs are:
Option A (La$$^{3+}$$, Ce$$^{4+}$$) and Option B (Yb$$^{2+}$$, Lu$$^{3+}$$).

Option A ( La$$^{3+}$$, Ce$$^{4+}$$ ), Option B ( Yb$$^{2+}$$, Lu$$^{3+}$$ )