Here C is:
 $ C{H_3}COOK\xrightarrow{{electrolysis}}A\xrightarrow[\Delta ]{{{{1000}^0}C}}\,B\xrightarrow[{alkaline}]{{KMn{O_4}\,}}C $
(A) $ C{H_3}COOH $
(B) $ C{H_2}OH - C{H_2}OH $
(C) $ CHO - CHO $
(D) $ COOH - COOH $

Hint : Electrolysis of carboxylic salts yields an alkane along with the liberation of carbon dioxide at the anode. This alkane on further decomposition at higher temperature yields an alkene and upon further oxidation of alkene, it will get converted into an alcohol.

Complete step by step answer
Decarboxylation is a reaction in which a carboxyl group is eliminated from the chain along with the liberation of carbon dioxide gas. It is mostly done in the reaction of carboxylic acids. It is a reversible process. You can produce alkane from the carboxylic acid by removing the $ C{O_2} $ . The Kolbe decarboxylation reaction involves the conversion of a carboxylic salt to an alkane with the liberation of $ C{O_2} $ at the anode. The salt could be a sodium salt or a potassium salt. In the given reaction, the salt is of potassium so when we treat it with $ {H_2}O $ . It will produce alkane with $ C{O_2} $ and $ NaOH $ and $ {H_2} $ . The reaction will be:
 $ C{H_3}COOK\xrightarrow[{{H_2}O}]{{electrolysis}}\,C{H_3} - C{H_3} + 2KOH + {H_2} + C{O_2} $
Hence $ A $ is Ethane. This ethane formed upon the catalytic cracking will get converted into an alkene at a high temperature. The reaction will be:
 $ C{H_3} - C{H_3}\xrightarrow[\Delta ]{{{{1000}^0}C}}C{H_2} - C{H_2} + {H_2} $
Hence $ B $ is ethane. This ethene produced in reaction with cold alkaline $ KMn{O_4} $ will undergo oxidation in which the double bonds between the carbon atoms will break and be replaced by an oxygen atom. In this way, alkene will get converted into ethylene glycol. This reaction occurs via the formation of an intermediate. The reaction will be;
 $ C{H_2} - C{H_2}\xrightarrow[{KMn{O_4}}]{{Alkaline}}C{H_2}OH - C{H_2}OH $
Hence $ C $ is $ C{H_2}OH - C{H_2}OH $ . Therefore option (B) is correct.

Note
Upon decarboxylation reaction, the carboxylic salt gets converted into alkane and when this alkane is treated at high temperature in the range of $ 750 - {1000^0}C $ , it gets converted into alkene and when you treat an alkene with alkaline potassium permanganate it will yield a diol.