The correct answer is A.
The reaction is a Friedel-Crafts alkylation, which is a reaction between an aromatic ring and an alkyl halide in the presence of a Lewis acid catalyst. The Lewis acid catalyst helps to activate the aromatic ring by polarizing the pi bond, making it more susceptible to attack by the alkyl halide.
In option A, the Lewis acid catalyst is $\ce{AlCl3}$. $\ce{AlCl3}$ is a strong Lewis acid and is therefore a good choice for this reaction. The alkyl halide is bromobenzene. Bromobenzene is a good alkyl halide to use in this reaction because it is a primary alkyl halide. Primary alkyl halides are more reactive than secondary or tertiary alkyl halides in Friedel-Crafts alkylations.
In option B, the Lewis acid catalyst is $\ce{FeCl3}$. $\ce{FeCl3}$ is a weaker Lewis acid than $\ce{AlCl3}$. Therefore, it is not as good a choice for this reaction. The alkyl halide is iodobenzene. Iodobenzene is a good alkyl halide to use in this reaction because it is a primary alkyl halide. However, iodobenzene is less reactive than bromobenzene in Friedel-Crafts alkylations.
In option C, the Lewis acid catalyst is $\ce{ZnCl2}$. $\ce{ZnCl2}$ is a very weak Lewis acid. Therefore, it is not a good choice for this reaction. The alkyl halide is chlorobenzene. Chlorobenzene is a good alkyl halide to use in this reaction because it is a primary alkyl halide. However, chlorobenzene is less reactive than bromobenzene or iodobenzene in Friedel-Crafts alkylations.
In option D, there is no Lewis acid catalyst. Therefore, this option is not a good choice for this reaction.
In conclusion, the most suitable reagent combination to bring out the following transformation is A.