The major product (P) in the reaction

The alkene reacted with $$\text{HBr}$$ acts as a nucleophile and attacks the proton ($$\text{H}^+$$). The double bond breaks to form the most stable possible carbocation intermediate.

The positive charge on the carbon directly attached to the benzene ring is a benzylic carbocation. It is exceptionally stable because the positive charge can be delocalized throughout the $$\pi$$-system of the aromatic ring via resonance.
Once the stable benzylic carbocation intermediate is formed, the bromide ion ($$\text{Br}^-$$) attacks the electron-deficient benzylic carbon.

Therefore, Option B is correct