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(C) The reaction involves the protonation of the hydroxyl group by $H^+$ from $Conc. \ HBr$,followed by the loss of water to form a stable carbocation

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(C) The reaction involves the protonation of the hydroxyl group by $H^+$ from $Conc. \ HBr$,followed by the loss of water to form a stable carbocation. The carbocation is stabilized by resonance with the adjacent double bond. The bromide ion $(Br^-)$ then attacks the electrophilic carbon to form the final product. The mechanism is as follows:$1$. Protonation of the $-OH$ group: $R-OH + H^+ \rightarrow R-OH_2^+$$2$. Loss of $H_2O$ to form a resonance-stabilized carbocation: $CH_3-CH(OH)-C(=CH_2)-COOCH_3 \rightarrow CH_3-CH^+-C(=CH_2)-COOCH_3 \leftrightarrow CH_3-CH=C(CH_2^+)-COOCH_3$$3$. Nucleophilic attack by $Br^-$: $CH_3-CH=C(CH_2^+)-COOCH_3 + Br^- \rightarrow CH_3-CH=C(CH_2Br)-COOCH_3$Thus,the major product is $CH_3-CH=C(CH_2Br)-COOCH_3$.

Consider the above reaction,the major product $P$ formed is:

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