Correct Option: Option 3 (which shows B as 4-bromostyrene, C as 1-bromo-4-(1-bromoethyl)benzene, and states that A and C are position isomers).

Let us analyze the reaction sequence step-by-step:

Step 1: Identify Reactant A
From the question image, the starting compound (Reactant A) is 1-bromo-4-(2-bromoethyl)benzene. It contains a benzene ring with a bromine atom ($-\mathrm{Br}$) at the para-position and a 2-bromoethyl group ($-{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$) at the opposite position.

Step 2: Elimination Reaction to form B
When Reactant A is treated with alcoholic sodium hydroxide (${\mathrm{NaOH}}_{\text{(alc)}}$), a strong base, it undergoes dehydrohalogenation (an E2 elimination reaction). The base abstracts a proton from the carbon beta to the leaving group on the side chain:
$-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{OH}}^{-}\to -\mathrm{CH}={\mathrm{CH}}_2+H_{2}O+{\mathrm{Br}}^{-}$
This elimination occurs on the alkyl side chain rather than the aromatic ring because aryl halides do not undergo elimination under these conditions. Thus, compound B is 4-bromostyrene (or 1-bromo-4-ethenylbenzene), which features a vinyl group ($-\mathrm{CH}={\mathrm{CH}}_2$) attached to the para-position of the bromobenzene ring.

Step 3: Hydrohalogenation Reaction to form C
When 4-bromostyrene (B) is treated with hydrobromic acid in ether ($\mathrm{HBr}/\text{Ether}$), it undergoes electrophilic addition across the carbon-carbon double bond. According to Markovnikov's rule, the electrophile ($H^{+}$) adds to the carbon with more hydrogen atoms (the terminal carbon) to generate the more stable carbocation intermediate:
$-\mathrm{CH}={\mathrm{CH}}_2+H^{+}\to -{\mathrm{CH}}^{+}-{\mathrm{CH}}_3$
This benzylic carbocation is highly resonance-stabilized by the aromatic ring. The nucleophile (${\mathrm{Br}}^{-}$) then attacks the carbocation to form the major product:
$-{\mathrm{CH}}^{+}-{\mathrm{CH}}_3+{\mathrm{Br}}^{-}\to -\mathrm{CH}\left(\mathrm{Br}\right)-{\mathrm{CH}}_3$
Therefore, compound C is 1-bromo-4-(1-bromoethyl)benzene.

Step 4: Relationship between A and C
Let us compare the structures of compound A and compound C:
- Compound A: 1-bromo-4-(2-bromoethyl)benzene
- Compound C: 1-bromo-4-(1-bromoethyl)benzene
Both compounds share the same molecular formula ($C_8{H}_8{\mathrm{Br}}_2$) but differ in the position of the bromine atom along the ethyl chain (carbon 2 in A versus carbon 1 in C). Therefore, compounds A and C are position isomers.