In the reaction given
 $\underset{\left( X \right)}{\mathop{{{C}_{3}}{{H}_{7}}Br}}\,\xrightarrow[\left( alco. \right)]{KOH}Y\xrightarrow{NBS}Z\xrightarrow{KCN}C{{H}_{2}}=CH-C{{H}_{2}}-CN$ X can be :
 (A) $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-Br$
 (B) $C{{H}_{2}}=CH\text{ }-C{{H}_{2}}-Br$
 (C) $C{{H}_{3}}-CH\text{ }\left( Br \right)-C{{H}_{3}}$
 (D) (a) and (c) both

Hint: Among the given options, option (A) and (C) are isomers of bromopropane. That is they are 1-bromopropane and 2-bromopropane respectively. When a haloalkane is treated with alcoholic KOH, the substrate undergoes dehydrohalogenation or elimination reaction.

Complete step by step answer:
 - Let’s start with the concept of isomerism. The compounds which have the similar molecular formula but having different arrangements of groups or atoms in space are called isomers and this occurrence is called isomerism.

 - The ${{C}_{3}}{{H}_{7}}Br$ or X is bromopropane and it can exist in both the forms as 1-bromopropane ($C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-Br$) and 2-bromopropane ($C{{H}_{3}}-CH\text{ }\left( Br \right)-C{{H}_{3}}$). They both are position isomers which have the same molecular formula ${{C}_{3}}{{H}_{7}}Br$ but different position of the bromine atom attached.
- The option (B) can already be eliminated since it contains only 5 hydrogen instead of 7 hydrogen in ${{C}_{3}}{{H}_{7}}Br$ . When a haloalkane such as bromopropane is treated with alcoholic KOH, the substrate undergoes dehydrohalogenation or elimination reaction and propene will be formed. The reaction can be shown below:
\[\underset{Bromopropane}{\mathop{{{C}_{3}}{{H}_{7}}Br}}\,\xrightarrow[{}]{Alc.KOH}\underset{prop-1-ene}{\mathop{C{{H}_{2}}=CH-C{{H}_{3}}}}\,\]

- This propene molecule will undergo reaction with N-Bromosuccinimide or NBS and the reaction is continued till the formation of But-2-enenitrile ( $C{{H}_{2}}=CH-C{{H}_{2}}-CN$).
From the above discussions it’s clear that the structure of X can be both (a) and (c) since they both will lead to the formation of propene by elimination reaction.
So, the correct answer is “Option D”.

Note: Keep in mind that in the given reaction elimination reaction takes place and in elimination reactions, the hydroxide ion (from alc.KOH) acts as a base and will remove hydrogen as a hydrogen ion from the carbon atom next to the one holding the bromine. The resultant re-arrangement of the electrons will expel bromine as a bromide ion and propene will be formed.