Bouveault-Blanc reduction reaction involves:
A. Reduction of a carbonyl compound with $Na/Hg$ and $HCl$
B. Reduction of an acyl halide with ${H_2}/Pd$
C. Reduction of an anhydride with $LiAl{H_4}$
D. Reduction of an ester with $Na/{C_2}{H_5}OH$

Hint:In order of the question, Bouveault-Blanc reduction reaction was used widely as before the availability of the reducing agent lithium aluminium hydroxide and other related reagents. This reaction particularly involves the use of alkali metal as a reagent.

Complete step-by-step answer: Bouveault-Blanc reaction was used particularly for the reduction of esters before the introduction of the reducing agent lithium aluminium hydroxide, which is now widely used for the reduction of esters.It reduces esters into alcohol in presence of alkali metal sodium and ethyl alcohol which gives the reaction products as terminal alcohol of that ester compound.The general schematic representation of the Bouveault-Blanc reduction reaction will go as shown below:
 $R - CO - O{R^1}\xrightarrow[{EtOH}]{{N{a^0}}}R - C - OH + {R^1}OH$
In the above reaction $N{a^0}$ acts as a single electron reducing agent and ethyl alcohol that is $EtOH$ acts as a proton donor.So, by analysing the above steps we have came a conclusion that Bouveault-Blanc reduction reaction involves the reduction of an ester with $Na/{C_2}{H_5}OH$ .

Additional Information: This reaction method is generally utilized as a cheap substitution for lithium aluminum hydroxide as decreases of esters in modern compound reactions. It was the lone elective strategy for decrease of esters which was being used before the development of the metal hydride reducing reagents.

Hence, the correct option is (D).

Note:In Bouveault-Blanc reduction reaction, while the reduction is proceeding if there is a proper source of the proton is not provided like ethyl alcohol is used in the above reaction then the dimerization of intermediate will take place instead of the alcohol product.