Step 1: Vicinal dibromide + $Zn/\Delta$ removes the two bromines to give a cyclopentene with the methyl group on the saturated carbon (3-methylcyclopent-1-ene).

Step 2: Addition of $HBr$ to the alkene proceeds via Markovnikov addition. The initial secondary carbocation undergoes a $1,2-H$ shift to form the more stable tertiary carbocation at the carbon bearing the methyl group.

Step 3: $B{r}^{-}$ attacks the tertiary carbon, placing the bromine on the same carbon as the methyl group.

$\therefore$ The major product (P) has $Br$ and $C{H}_3$ on the same ring carbon, matching option (2).