The major product of the following reaction is:

In the presence of organic peroxide, the reagent $$HBr$$ adds to an alkene by a free-radical (anti-Markovnikov) mechanism, not by the normal ionic (Markovnikov) route.

Given alkene: $$CH_3CH=CH_2$$

Step 1 - Initiation
$$ROOR \;\xrightarrow{\;\Delta\;/\;h\nu\;} 2\,RO\cdot$$ (alkoxy radicals)

Step 2 - Generation of bromine radical
$$RO\cdot + HBr \rightarrow ROH + Br\cdot$$

Step 3 - Propagation (anti-Markovnikov addition)
Br-radical attacks the less substituted end of the double bond to give the more stable secondary carbon radical:
$$Br\cdot + CH_3CH=CH_2 \rightarrow CH_3\dot{C}HCH_2Br$$

The carbon radical then abstracts H• from another $$HBr$$ molecule:
$$CH_3\dot{C}HCH_2Br + HBr \rightarrow CH_3CH_2CH_2Br + Br\cdot$$

The chain continues until radicals recombine and terminate. The net result is addition of $$H$$ to the more substituted carbon and $$Br$$ to the terminal carbon—exactly opposite to Markovnikov’s rule.

Thus the major product is $$CH_3CH_2CH_2Br$$ (1-bromopropane).

Option B which is: 1-bromopropane.