Answer
The most appropriate drawing for that situation is the number $(iv)$
Work Step by Step
At the equivalence point, the $HA$ was totally neutralized by the $NaOH$, and the solution is formed by: $Na^+$ ions, $H_2O$ molecules (both ommited by the drawing) and $A^-$ ions.
$HA(aq) + NaOH(aq) -- \gt A^-(aq) + Na^+(aq) + H_2O(l)$
Therefore, the best representation is in the image number $(iv)$
