To solve this problem, we need to understand the reaction of an alkene with HBr.$\\$The given compound is $C_6H_5CH=CHCH_3. \\$This is an example of an electrophilic addition reaction, following Markovnikov's rule.$\\$Markovnikov's rule states that the hydrogen atom from HBr will add to the carbon with more hydrogen atoms,$\\$and the bromine atom will add to the carbon with fewer hydrogen atoms.$\\$Step-by-step reaction:$\\$1. The double bond between $CH=CH$ is the site of reaction.$\\$2. The $CH_3$ group is attached to the second carbon of the double bond.$\\$3. According to Markovnikov's rule, the H from HBr will add to the $CH$ group,$\\$ and Br will add to the $CH_2$ group.$\\$4. The resulting product is $C_6H_5CHBrCH_2CH_3. \\$This corresponds to Option 4.$\\$Therefore, the correct product of the reaction is $C_6H_5CHBrCH_2CH_3.$