For $KCl,NaCl,LiCl$ anionic part is same.$Li^+$(aq) has smallest size means high charge density .So it has high hydration and hence low ionic mobility .So it has low molar conductance .

So $\Lambda_m^{\infty}KCl > \Lambda_m^{\infty}NaCl > \Lambda_m^{\infty}LiCl$

Hence (a) is the correct answer.