3-Phenylpropene on reaction with $\mathrm{HBr}$ gives (as a major product) 

According to Markownikoffs rule. the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond. 

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3+\mathrm{HBr}\to {\mathrm{C}}_6{\mathrm{H}}_5\stackrel{ }{\mathrm{CH}\left(Br\right){\mathrm{CH}}_2}{\mathrm{CH}}_3$