When ${\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{CH}$ reacts with sodium an intermediate ${\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{CNa}$ is formed which on further reaction with propyl bromide gives ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{C}\equiv {\mathrm{CCH}}_3$ as the product. The second step of the reaction follows, ${\mathrm{S}}_{\mathrm{N}}2$ mechanism.

The reaction is as follows,

Hence in the above reaction,

$\mathrm{A}={\mathrm{CH}}_3-\mathrm{C}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+},\mathrm{B}={\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Br}$

Hence option A is the answer.