Electrophilic addition of $HBr$ to an unsymmetrical alkene follows Markovnikov's rule, which states that the hydrogen of the adding molecule attaches to the doubly bonded carbon that already carries more hydrogen atoms. The mechanism explains why: the proton adds first to give the more stable carbocation. With propene, $C{H}_3-CH=C{H}_2$, protonation at the terminal carbon produces a secondary carbocation $C{H}_3-C{H}^{+}-C{H}_3$, which is stabilized by hyperconjugation and the inductive effect of two alkyl groups, whereas protonation at the internal carbon would give a less stable primary carbocation. Bromide then attacks the secondary cation to give 2-bromopropane as the major product. 1-Bromopropane is the anti-Markovnikov product that forms only in the presence of peroxide, so it is wrong here. 1,2-Dibromopropane would require addition of $B{r}_2$, not $HBr$, so it is incorrect. Propan-2-ol involves water, not bromide, so it does not apply. This is the classic NCERT illustration of Markovnikov addition. A plausibility check: the dominant route always passes through the more stable carbocation, confirming the secondary bromide.