Why $H(A)=H(C)$ where $C$ is $A$ with an additional parity bit?
Because the entropy represents information quantity, or if being measured in bit, the smallest number of bits per symbol we need to represent a source.

The source $A$ contains $4$ equiprobable symbols hence it is obvious that we need $\log_2(4) = 2$ bits per symbol to represent the source.

The source $C$ is simply $A$ with its parity bits hence no new information is added. We have more bits per symbols, but the number of information bits is unchanged.

In communications, these additional bits are added to scope with error of transmissions.