Reagent selectivity. LiBH${}_4$ selectively reduces the ester (–CO${}_2$Et $\to$ –CH${}_2$OH) and leaves the carboxylic acid intact. BH${}_3$ behaves oppositely on this substrate: it preferentially reduces the carboxylic acid (–CO${}_2$H $\to$ –CH${}_2$OH) and leaves the ester intact.

Lactonisation under acidic workup. In each case the resulting hydroxy-acid (or hydroxy-ester) undergoes intramolecular cyclisation to form a five-membered lactone. The lactone is formed with retention of the existing stereocentres on the ring carbon.

Top substrate. Starting from one diastereomer:

• With LiBH${}_4$ then H${}^{+}$: ester reduces to CH${}_2$OH, then the CH${}_2$OH cyclises onto the carboxylic acid, giving lactone $\mathbf{P}$.
• With BH${}_3$ then H${}^{+}$: carboxylic acid reduces to CH${}_2$OH, then the CH${}_2$OH cyclises onto the ester, giving lactone $\mathbf{Q}$.

$\mathbf{P}$ and $\mathbf{Q}$ both contain the same skeleton but the relative orientation of the CH${}_3$ substituent versus the lactone oxygen/carbonyl differs because the new CH${}_2$O and C=O attachments are swapped at the ring carbon. They are therefore diastereomers.

Bottom substrate (opposite stereochemistry of the substituted ring carbon). The same argument applies: $\mathbf{R}$ and $\mathbf{S}$ are also diastereomers, differing in which of the two CH${}_2$O / C=O ends is attached at the ring carbon.

Hence $\mathbf{P}$ & $\mathbf{Q}$ are diastereomers and $\mathbf{R}$ & $\mathbf{S}$ are diastereomers. Option (C).