Reduction of epoxide
The correct question (taken from the 2016 JEE Paper is):

Lithium Aluminium Hydride reduces epoxides1, hence (A) would not give the required product. LAH also reduces ethers to primary alcohols, which is an easier way of eliminating the option (A). $\ce{BH3}$ in THF can reduce the acid, as explained in this question, hence the requisite product would not be obtained.
$\ce{NaBH4}$ cannot reduce esters, acids or epoxides in the absence of catalysts such as $\ce{PdCl2}$ or $\ce{AlCl3}$ and diglyme, hence (C) would be a correct option here.
Can you tell what Raney Ni would do here?
Raney Ni Reduces carboxylic acid and esters at elevated temperature (eg: $\pu{350 K}$); and no reaction with epoxide. So, it can be used for selective reduction of aldehyde at room temperature ($\pu{298 K}$).
Use this resource to help yourself out and as a reference for such reactions in the future.
References:

Murphy, David K., et al. “Reduction of Epoxides. III. Lithium Aluminum Hydride and Mixed Hydride Reduction of Some Secondary-Tertiary Epoxides.” Journal of the American Chemical Society, vol. 91, no. 10, May 1969, pp. 2649–53. doi:10.1021/ja01038a042.