When CH3C≡CH reacts with sodium an intermediate CH3C≡CNa is formed which on further reaction with propyl bromide gives CH3CH2CH2C≡CCH3 as the product. The second step of the reaction follows, SN2 mechanism.
The reaction is as follows,

Hence in the above reaction,
A=CH3−C≡C−Na+,B=CH3−CH2−CH2−Br
Hence option A is the answer.