1. We need a reagent that converts benzoic acid $C_6H_5COOH$ into benzaldehyde $C_6H_5CHO$ This is a partial reduction of a carboxylic acid to an aldehyde.

2. Now check each option:

Option A: $LiAlH_4$

$LiAlH_4$ is a very strong reducing agent. It reduces carboxylic acids all the way to primary alcohols.

So, $C_6H_5COOH \xrightarrow{LiAlH_4} C_6H_5CH_2OH$ This gives benzyl alcohol, not benzaldehyde.

So, A is incorrect.

Option B: $KMnO_4$

$KMnO_4$ is a strong oxidizing agent, not a reducing agent. It cannot convert benzoic acid to benzaldehyde.

So, B is incorrect.

Option C: MnO

Manganese oxide reagents are known for oxidation of alcohols, but in standard JEE chemistry, MnO does not directly convert benzoic acid to benzaldehyde in one step. There is no standard one-step reduction of benzoic acid to benzaldehyde using MnO.

So, C is not correct.

Option D: $NaBH_4$

$NaBH_4$ is a mild reducing agent. It reduces aldehydes and ketones, but it generally does not reduce carboxylic acids.

So, D is incorrect.

3. Therefore, among the given options, none is a proper reagent for converting benzoic acid directly to benzaldehyde in one step.

A common valid method is conversion of acid chloride followed by Rosenmund reduction, but that is not one of the options.

Hence, the stored answer appears to be incorrect.