Benzoic acid on reduction with di-isobutyl aluminium hydride (DIBALH) gives benzaldehyde.
$\displaystyle PhCOOH \xrightarrow {DIBALH} PhCHO $
Di-isobutyl aluminium hydride (DIBALH) is a reducing agent. It reduces carboxylic acids to aldehydes.

$-COOH$ can be converted into $CH _3$ group by heating with a mixture of HI and red phosphorus.
$CH _3COOH  \xrightarrow {\;Hl\;&\;red\;phosphorous} CH _3-CH _3$

$CH _{3}COOH\xrightarrow[]{LiAlH _{4}}C _{2}H _{5}OH$.
$ LiAlH _{4} $ being a strong reducing agent can reduce acids to alcohols.

$LiAlH _{4}$ is a very strong reducing agent. It can reduce acids directly to alcohols.

D

B
$LiAlH _{4}$ reduces ethanoic acid to ethanol and then $H _{2}$ / Ni reduces it further to alkane.

$BH _3$ in $THF$ being a strong reducing agent, can efficiently convert carboxylic acid into the corresponding primary alcohol. Also, $LiAlH _4$ being a strong reducing agent, releases $H^-$ ions in enough quantity to convert carboxylic acid to the corresponding alcohol. So, ethanoic acid can be converted to ethanol using $BH _3$ in $THF$ and $LiAlH _4$.

$CH _3OH+CO\xrightarrow [ (High\quad Pressure) ]{ CO/Rh } \underbrace{ CH _3COOH} _{Compound A}$

$RCOOH  +  LiAlH _4 \rightarrow RCHO  \rightarrow RCH _2OH $

using LAH (lithium aluminium hydride), the COOH group is reduced to  $ CH _{2} OH$  group

$RCOOH + 4H \xrightarrow{LiAlH _4}RCH _2OH+H _2O$

Reaction:
$CH _3-(CH _2) _{14}-COOH \overset{LiAlH _4}{\longrightarrow} \,CH _3-(CH _2) _{14}-CH _2OH \overset{HCl}{\longrightarrow} CH _3-(CH _2) _{14}-CH _2Cl$
$ CH _3-(CH _2) _{14}-CH _2Cl \overset{1.\, Mg/Et _2O}{\underset{2.\, Oxirane}{\longrightarrow}} CH _3-(CH _2) _{16}-CH _2OH \overset{KMnO _4}{\longrightarrow} CH _3-(CH _2) _{16}-COOH$