Reduction of an aldehyde with $\mathrm{NaBH_4}$ or $\mathrm{LiAlH_4}$ produces:

Aldehydes ($\mathrm{R-CHO}$) are reduced to **primary** alcohols ($\mathrm{R-CH_2OH}$) because the carbonyl carbon ends up with just one $\mathrm{R}$ group on it. Ketones ($\mathrm{R_2C=O}$) reduce to *secondary* alcohols (option B) for the symmetric reason. Neither $\mathrm{NaBH_4}$ nor $\mathrm{LiAlH_4}$ goes all the way to alkane — that requires harsher conditions (Clemmensen, Wolff–Kishner).