The product (A) formed in the following reaction sequence is

The given sequence converts an alkyl halide into an alkyne by the well-known elimination → addition → double-elimination → alkylation route. For clarity, let us denote the starting compound as $$CH_3CH_2Br$$ (ethyl bromide) and follow every step.

Step 1 : β-elimination (dehydrohalogenation)
Reagent : alcoholic $$KOH$$ (strong base).
Reaction : $$CH_3CH_2Br \xrightarrow[\text{alc.}]{KOH,\;\Delta} CH_2\!=\!CH_2 + KBr + H_2O$$
Product : ethene.

Step 2 : Electrophilic addition of bromine
Reagent : $$Br_2/CCl_4$$ (cold, non-polar solvent).
Reaction : $$CH_2\!=\!CH_2 + Br_2 \longrightarrow CH_2Br-CH_2Br$$ (1,2-dibromoethane).
Thus we have introduced two vicinal bromine atoms, which are a precursor for preparing an alkyne.

Step 3 : Double dehydrohalogenation
Reagent : excess alcoholic $$KOH$$ (or $$NaNH_2$$, any strong base).
Each $$KOH$$ removes one molecule of $$HBr$$; two such eliminations convert the vic-dibromide into an alkyne:
$$CH_2Br-CH_2Br \xrightarrow[\text{alc.}]{2\,KOH,\;\Delta} HC\!\equiv\!CH + 2\,KBr + 2\,H_2O$$
Product : ethyne (acetylene), a terminal alkyne having an acidic hydrogen.

Step 4 : Formation of acetylide ion
Reagent : $$NaNH_2$$ (sodamide, very strong base).
Reaction : $$HC\!\equiv\!CH + NaNH_2 \longrightarrow HC\!\equiv\!C^{-}Na^{+} + NH_3$$
The terminal proton is removed, giving the nucleophilic acetylide ion.

Step 5 : Alkylation (SN2) with ethyl bromide
Reagent : another mole of $$CH_3CH_2Br$$.
The acetylide ion undergoes an $$S_N2$$ reaction with the primary alkyl bromide:
$$HC\!\equiv\!C^{-}Na^{+} + CH_3CH_2Br \longrightarrow CH_3CH_2C\!\equiv\!CH + NaBr$$

Final product (A) : $$CH_3CH_2C\!\equiv\!CH$$, i.e. 1-butyne.

Among the given choices, Option B corresponds to 1-butyne, so

Option B which is: 1-butyne.