Let's assume the unit digit of the number is $y$ & $x$ is the digit of ten's place.
∴ The number is $10x+y$
Now, given $x+y =12$
if the number is $yx$, then $\{(10y+x)-(10x+y)\}=54$
Now, we know, $x+y =12$
∴ $y=12-x$
Now, $\{(10y+x)-(10x+y)\}=54$
$10y+x-10x-y=54$
$10(12-x)+x-10x-(12-x)=54$
Or,$120-10x+x-10x-12+x=54$
Or, $120-20x+2x-12=54$
Or,$108-18x=54$
Or,$108-54=18x$
Or,$18x=54$
Or,$x=3$
∴$ y = 12-x = 12-3 = 9$
The original number is $39$ which is option A)