${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{C}\equiv \mathrm{N}\stackrel{ {\mathrm{H}}_2\mathrm{O}/{\mathrm{H}}^{+}}{\to }\mathrm{A}\left(\mathrm{final}\right) \stackrel{ {\mathrm{Br}}_2/\mathrm{Red} \mathrm{P}}{\to }\mathrm{B} \stackrel{ \mathrm{aq}. \mathrm{NaOH}}{\to }\mathrm{C}$

The incorrect statement regarding ‘C’ is 

From the reaction sequence, the product C formed is Lactic acid also known as 2-Hydroxypropanoic acid. And the incorrect statement about 2-Hydroxypropanoic acid (product C) is that it is a meso compound. Because achiral compounds having multiple chiral centers which are optically inactive are called meso compounds. 

Here, in compound C, there is only one chiral center and is said to show optical activity, so it’s not a meso compound.

Therefore, option (D) is correct.