In binary subtraction, we find the 2's complement of operand $b$ by inverting all bits and adding a $1$. When adding this to $a$ you have performed $a-b$. In BCD we have to find the 10's complement. This is done by subtracting 9 from each decimal digit (a 4 bit binary number) and adding a $1$. Now, using ordinary BCD addition rules we add the two numbers.
Example:
10's complement of 0101 0110 is 0100 0100 (subtracted 9 from each 4 bit segment and added a $1$ to the last one. Notice that this could overflow when
finding the 10's complement of $0000$. You would have to correct the BCD number.).
Now adding 1001 0011 and 0100 0100:
$1001 0011 + 0100 0100=1101 0111$
Correction is done when a digit acceeds the value 9 ($1001$ in BCD).
The first digit is correct.
The second digit is invalid. To correct it we add 6 ($0110$ in BCD). $1101+0110=0011$ (ignore overflow).
In binary subtraction, we find the 2's complement of operand $b$ by inverting all bits and adding a $1$. When adding this to $a$ you have performed $a-b$. In BCD we have to find the 10's complement. This is done by subtracting 9 from each decimal digit (a 4 bit binary number) and adding a $1$. Now, using ordinary BCD addition rules we add the two numbers.
Example:
10's complement of 0101 0110 is 0100 0100 (subtracted 9 from each 4 bit segment and added a $1$ to the last one. Notice that this could overflow when finding the 10's complement of $0000$. You would have to correct the BCD number.).
Now adding 1001 0011 and 0100 0100:
$1001 0011 + 0100 0100=1101 0111$
Correction is done when a digit acceeds the value 9 ($1001$ in BCD). The first digit is correct. The second digit is invalid. To correct it we add 6 ($0110$ in BCD). $1101+0110=0011$ (ignore overflow).
So:
$1101 0111$ becomes $0011 0111$
Therefore:
$1001 0011 - 0101 0110=0011 0111$
Let's check if this is correct:
$1001 0011=93$
$0101 0110=56$
$93-56=37=0011 0111$