Step 1: Set up the 4f count using $n=Z-57$ for $\mathrm{Ln^{3+}}$.
$\mathrm{Nd^{3+}}(Z=60)\to 4f^3$, $\mathrm{Pm^{3+}}(Z=61)\to 4f^4$, $\mathrm{Ho^{3+}}(Z=67)\to 4f^{10}$, $\mathrm{Er^{3+}}(Z=68)\to 4f^{11}$.
Step 2: Use electron-hole pairing to shortcut $S$ and $L$.
The 4f shell is half full at $4f^7$. A configuration $4f^{7+k}$ has the same $S$ and $L$ as $4f^{7-k}$, because both have the same number of unpaired electrons filling the $m_l$ slots from the outside in.
$4f^{10}=4f^{7+3}$ pairs with $4f^4=4f^{7-3}$, so $\mathrm{Ho^{3+}}$ shares $S,L$ with $\mathrm{Pm^{3+}}$.
$4f^{11}=4f^{7+4}$ pairs with $4f^3=4f^{7-4}$, so $\mathrm{Er^{3+}}$ shares $S,L$ with $\mathrm{Nd^{3+}}$.
This already flags the two candidate matching pairs: $\{\mathrm{Pm^{3+}},\mathrm{Ho^{3+}}\}$ and $\{\mathrm{Nd^{3+}},\mathrm{Er^{3+}}\}$.
Step 3: Confirm $S$ and $L$ for the base configurations.
$4f^3$: unpaired electrons in $m_l=3,2,1$ give $S=3/2$, $M_L=6$, so $L=6$, term ${}^4I$.
$4f^4$: unpaired electrons in $m_l=3,2,1,0$ give $S=2$, $M_L=6$, so $L=6$, term ${}^5I$.
Step 4: Apply Hund's third rule for $J$, since the pairs only match in $S,L$, not $J$.
$\mathrm{Nd^{3+}}$ ($4f^3$, less than half filled): $J=L-S=6-1.5=9/2$, term ${}^4I_{9/2}$.
$\mathrm{Er^{3+}}$ ($4f^{11}$, more than half filled): $J=L+S=6+1.5=15/2$, term ${}^4I_{15/2}$. These differ in $J$, confirming pair (B).
$\mathrm{Pm^{3+}}$ ($4f^4$, less than half filled): $J=L-S=6-2=4$, term ${}^5I_4$.
$\mathrm{Ho^{3+}}$ ($4f^{10}$, more than half filled): $J=L+S=6+2=8$, term ${}^5I_8$. These differ in $J$, confirming pair (A).
Step 5: Rule out the cross pairs.
$\mathrm{Pm^{3+}}$ and $\mathrm{Nd^{3+}}$ come from different base configurations ($4f^4$ vs $4f^3$), so their $2S+1$ (5 vs 4) do not match; the same applies to $\mathrm{Ho^{3+}}$ and $\mathrm{Er^{3+}}$.
Final Answer:
The hole-pairing shortcut pins down the two matching pairs directly: $\mathrm{Pm^{3+}}$/$\mathrm{Ho^{3+}}$ and $\mathrm{Nd^{3+}}$/$\mathrm{Er^{3+}}$.
\[ \boxed{\text{(A) and (B)}} \]