Considering LiBH_4 reduces an ester group to the corresponding...

From the given statement, $LiBH_4$ selectively reduces the ester group ($-CO_2Et$) to an alcohol ($-CH_2OH$) and leaves the carboxylic acid ($-CO_2H$) unchanged.

Borane ($BH_3$), on the other hand, is a well-known reagent for the chemoselective reduction of carboxylic acids to alcohols in the presence of esters.

Let us analyze the first set of reactions yielding P and Q:
The starting material is a 1,2-disubstituted cyclobutane derivative. At C1, the ester group ($-CO_2Et$) is on a wedge (up) and the acid group ($-CO_2H$) is on a dash (down). The adjacent C2 has a methyl group.
In product P (using $LiBH_4$), the wedge $-CO_2Et$ is reduced. Thus, the new $-CH_2OH$ group is on the wedge.
In product Q (using $BH_3$), the dash $-CO_2H$ is reduced. Thus, the new $-CH_2OH$ group is on the dash.
Comparing P and Q, the stereocenter at C1 is inverted relative to the unchanged stereocenter at C2. Since their relative stereochemistry (cis vs trans relationship between $-CH_2OH$ and $-CH_3$) is different, P and Q are diastereomers.

Now let us analyze the second set of reactions yielding R and S:
The starting material is a 1,3-disubstituted cyclobutane derivative. At C1, $-CO_2Et$ is on a wedge and $-CO_2H$ is on a dash. At C3, there is a methyl group.
In product R (using $LiBH_4$), the ester on the wedge is reduced. The resulting $-CH_2OH$ group is on the wedge (cis to the methyl group if we assume methyl is up).
In product S (using $BH_3$), the acid on the dash is reduced. The resulting $-CH_2OH$ group is on the dash (trans to the methyl group).
Since R and S represent the cis and trans isomers of a 1,3-disubstituted cycloalkane, they are stereoisomers that are not mirror images. Therefore, R and S are also diastereomers.

Thus, P & Q are diastereomers, and R & S are diastereomers.