Answer
The most appropriate representation is in the drawing number $(ii)$.
Work Step by Step
After the equivalence point, there is an excess of $NaOH$ molecules, therefore, the best representation is the drawing number $(ii)$
At the equivalence point, the $HA$ was totally neutralized by the $NaOH$, therefore, after that point, if we add more base, it will be an excess of the compound, and the image number $(ii)$ has these $OH^-$, which are only visible after the equivalence point.
