How to identify the asymmetric carbon in [4-(1-bromoethylidene)cyclohexyl]acetaldehyde?
There is probably little reason in discussing the hydrogen and the $\ce{CH2CHO}$ group which are definitely different residues. The question remains whether the sides of the ring are identical or not.

If you go along the top side of the ring, you will encounter:


$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CMeBr}$


Looking into it a little deeper, we see that we are coming from the (Z) side of the double bond.

Going along the bottom side you will encounter:


$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CBrMe}$


Here, however, we are arriving from the (E) side of the double bond.

Since (E) is not equal to (Z) the two sides of the ring are not equal, therefore the carbon has four different residues, therefore it is chiral.