In the reaction 
$\text{H−C≡CH}\frac{\left(1\right){\mathrm{NaNH}}_2/ \mathrm{liq}. {\mathrm{NH}}_3}{\left(2\right){\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}}\mathrm{X}\frac{ (1) {\mathrm{NaNH}}_2/ \mathrm{liq}. {\mathrm{NH}}_3}{\left(2\right){\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}}\mathrm{Y}$  X and Y are 

Given Reaction:

H−C≡CH → This is ethyne (acetylene)

Reagents:

NaNH₂ / liq. NH₃ → strong base, forms acetylide ion

Then reacts with CH₃CH₂Br (ethyl bromide) → alkylation (adds ethyl group)

Step-by-step:
Step 1:

Ethyne (H−C≡CH) + NaNH₂
→ forms acetylide ion (⁻C≡C⁻ Na⁺)

Step 2:

This acetylide ion reacts with ethyl bromide (CH₃CH₂Br)
→ substitutes Br and forms Butyne

So:

CH≡CH + CH₃CH₂Br → CH≡C−CH₂CH₃

This is 1-butyne
So, X = 1-Butyne

Step 3:

Now take 1-Butyne (CH≡C−CH₂CH₃)
Again treat with NaNH₂ / liq. NH₃
→ again forms a new acetylide ion at terminal H

Then react it again with CH₃CH₂Br

Now:

CH≡C−CH₂CH₃ + CH₃CH₂Br → CH≡C−CH₂CH₂CH₂CH₃

This is 2-Hexyne (triple bond at position 2)

So, Y = 2-Hexyne

Final Answer:

X = 1-Butyne; Y = 2-Hexyne
Correct Option: (a)