Arrange the carbanions in decreasing order of stability: (P) $\mathrm{HC}\equiv \mathrm{C}{}^{-}$, (Q) $\mathrm{CH}{}_2=\mathrm{CH}{}^{-}$, (R) $\mathrm{CH}{}_3\mathrm{CH}{}_2{}^{-}$, (S) $(\mathrm{CH}{}_3){}_2\mathrm{CH}{}^{-}$.

The lone pair of a carbanion is more stable in an orbital with greater s character (sp > sp2 > sp3) because the carbon is effectively more electronegative. Among sp3 carbanions, the +I effect of the extra methyl group makes the secondary one less stable than the primary one. So P > Q > R > S.