$( A ) \rightarrow( B )$ Free $H ^{+}$ions are replaced by $Na ^{\oplus}$ which decreases conductance.
$(B) \rightarrow (C)$ Un-dissociated benzoic acid reacts with $NaOH$ and forms salt which increases ions \ conductance increases.
$(C) \rightarrow (D)$ After equivalence point at (3), $NaOH$ added further increases $Na ^{\oplus}$ and $OH ^{\circ}$ ions which further increases the conductance.
$(a)$ $ 10^{-10}$ $(b)$ $\frac{{Kw}}{{{{10}^{ - 10}}}}$ $(c)$ $\frac{{Kw}}{{{{10}^{ - 8}}}}$ $(d)$ $10^{-4}$