Use the sodalime reaction first

Heating a carboxylic acid salt with sodalime causes decarboxylation: $RCOONa \xrightarrow[\Delta]{\text{sodalime}} RH$

Since [P] on heating with sodalime gives toluene $(C_6H_5CH_3)$, [P] must be a methyl benzoic acid (toluic acid), because removing $-COOH$ from a methyl-substituted benzoic acid gives toluene.

So [P] is one of:

• $o$-methyl benzoic acid ($o$-toluic acid)
• $m$-methyl benzoic acid ($m$-toluic acid)
• $p$-methyl benzoic acid ($p$-toluic acid)

Now analyze bromination with $Br_2/FeBr_3$ in $CCl_4$

This is electrophilic aromatic substitution.

In methyl benzoic acids, the substituents have directing effects:

• $-CH_3$ is ortho/para directing and activating.
• $-COOH$ is meta directing and deactivating.

We need the compound that gives a single monobromo isomer.

Check each possible isomer

(i) $o$-methyl benzoic acid: $2$-methyl benzoic acid

Positions available for bromination are not all equivalent, so more than one product is possible. Hence, not suitable.

(ii) $m$-methyl benzoic acid: $3$-methyl benzoic acid

Again, the free positions are not symmetry equivalent, so bromination can give more than one isomer. Hence, not suitable.

(iii) $p$-methyl benzoic acid: $4$-methyl benzoic acid

Structure has para-substituted groups:

• $-CH_3$ at position 1
• $-COOH$ at position 4

The unsubstituted positions are 2, 3, 5, 6. By symmetry:

• positions 2 and 6 are equivalent
• positions 3 and 5 are equivalent

Now directing effects:

• $-CH_3$ directs to 2, 6
• $-COOH$ directs to 2, 6 (meta to $-COOH$)

So both groups direct bromination to the same positions (2 and 6), and these two are equivalent by symmetry.

Therefore, only one monobromo product is formed.

Verify molecular formula

$p$-methyl benzoic acid has formula: $C_8H_8O_2$ Replacing one ring hydrogen by bromine gives: $C_8H_7O_2Br$ which matches the given product formula.

Conclusion

Hence [P] is $p$-methyl benzoic acid ($p$-toluic acid, 4-methyl benzoic acid).

Therefore, the correct option is C.