Consider the above sequence of reactions. The number of bromine...

The reaction sequence is as follows:
1. Bromination of nitrobenzene with $Br_2/FeBr_3$ results in meta-bromonitrobenzene because the $-NO_2$ group is meta-directing. (1 Br atom)
2. Reduction of the nitro group using $Sn/HCl$ followed by pH neutralization converts $-NO_2$ to $-NH_2$. The product is m-bromoaniline.
3. Reaction with $Br_2/H_2O$ (bromine water) on m-bromoaniline: The $-NH_2$ group is highly activating and ortho/para directing. In m-bromoaniline, the positions ortho and para to $-NH_2$ are vacant (positions 2, 4, and 6 relative to $-NH_2$). Bromine water causes polybromination at all available ortho and para positions. This adds 3 more bromine atoms. (Total 1 + 3 = 4 Br atoms)
4. Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts the $-NH_2$ group into a diazonium salt $-N_2^+Br^-$.
5. Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a bromine atom. (Total 4 + 1 = 5 Br atoms)
The final product (P) is 1,2,3,4,5-pentabromobenzene. Therefore, the number of bromine atoms in the final product is 5.