Q. Give reason why does this happen

A : $CH \equiv C-C{H}_{2_{ }}-CH=C{H}_2 add up HBr to give$

$CH \equiv C-C{H}_{2_{ }}-CH-C{H}_{3 }while CH \equiv C-CH = C{H}_2$
                  |
               Br
adds up HBr to give $C{H}_2 \equiv C-C{H}_{{}_{ }}= C{H}_2$
                                              |
                                            Br

Hi there,
The answer is quite simple.
For an electrophilic addition reaction in very 1st step electrophile attack so carbocation is formed as a intermediate.
In alkene carbocation with single bond is formed which has more no.of alpha hydrogen adding to its stability due to more hyperconjugation and as you know more stable carbication so more reactive alkene or alkyne for electrophilic addition reaction.
 
In alkyne carbocation with = bond is formed so it has less alpha hydrogen so less stability of carbocation so less reactive
Hence, alkynes are less reactive than alkenes and hence in first case, double bond cleaves, and HBr is added according to markovnikovs rule. But in 2nd reaction, the triple and double bond is seperated by a single bond. Such configuration is called conjugation as you see in benzene. Hence, the compound undergoes conjugation and double bond is unavailable for bonding. Hence HBr addition takes place on triple bond. Refer images for more details.

Note, in all general cases, 1st reaction only occurs, ie double bond undergoes addition. In these special cases, the addition occurs on triple bonds