The major product $\left(\mathrm{P}\right)$ formed in the below reaction is $\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2+$ Br $(1$ mole $)+{\mathrm{CCl}}_3$ at $253\mathrm{ }\mathrm{K}\to$

According to the reactivity of bonds, double bond, = is more reactive than triple bond, . In the given reactant, the bromine will attack the double bond carbon.

So, $\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2\stackrel{ }{+\mathrm{Br}\stackrel{{\mathrm{CCl}}_3 \mathrm{at} 253\mathrm{K}}{\to }}\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-\mathrm{CHBr}-{\mathrm{CH}}_2\left(\mathrm{Br}\right)$

Hence, the major product is $\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-\mathrm{CH}\left(\mathrm{Br}\right)-{\mathrm{CH}}_2\left(\mathrm{Br}\right)$.