I know that A is the product of a Knoevenagel condensation between the aldehyde and the diethyl malonate, due to the acidic protons of the latter, but I don't know what the reaction between that product and potassium cyanide would be.
Here is what I think is going on:
$Step$ $1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate.
$Step$ $2$ is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B
$Step$ $3$ is loss of one carboxy group by acid catalysed decarboxylation to give the cinnamic acid C
$Step$ $4$ formation of the mixed anhydride D
$Step$ $5$ cyclisation through formation of an aluminium-acyl complex literature example here to give the final product 6-cyclohexyl-Inden-1-one E
@Waylander has provided a well-reasoned solution to the post. I offer a different interpretation. Knoevenagel product 1a has been converted in high yield to cyano ester 2a and subsequently hydrolyzed to phenylsuccinic acid (3a) 1. In the addition of cyanide to the double bond of 1a and protonation, one of the labile ester groups is cleaved by cyanide. (e.g., Diethyl malonate is readily monosaponified by base in the cold.) Cyclization of anhydride 4a has been accomplished by standard Friedel-Crafts methods but, like Waylander, I am a big fan of polyphosphoric acid (PPA). The cyclohexyl series b (R = c-C6H11) is likely to lead to ketoacid 5b.
1) C. F. H. Allen and H. B. Johnson, Organic Synthesis, Coll. Vol. 4, 804, 1963.; Vol. 30, 83, 1950.