What does "2 digit BCD number" mean?

The idea here is that you can treat a BCD digit as a single entity. Look at the components you have:

• 1-digit adder
• Multiplexer (presumably for BCD digits)
• 9's complement unit

All of these use full BCD digits for their inputs and outputs, so you don't have to think about the individual bits. \$C\$ and \$ADD\$ are binary, but they never combine with the BCD signals -- they would only be used for the multiplexer selection signals.

So your circuit's inputs are:

• \$C\$, a binary signal
• \$ADD\$, a binary signal
• \$A\$, a two-digit BCD number with digits \$A_0\$ and \$A_1\$
• \$B\$, a two-digit BCD number with digits \$B_0\$ and \$B_1\$

And the outputs are:

• \$S\$, a collection of four BCD digits, which could be either:
  • A single number that represents the sum or difference of the inputs, with digits \$S_0\$ through \$S_4\$.
  • Two numbers which represent the complements of the two inputs, with \$S_1S_0 = \mathrm{complement}(A_1A_0)\$ and \$S_3S_2 = \mathrm{complement}(B_1B_0)\$.

One 2-digit BCD number is 8-bit wide and is made from

• High nibble = \$A_1\$ and

• Low nibble = \$A_0\$

Each nibble consists of 4 bits.
So e.g. \$A_1 = 1000, A_0 = 0110\$ gives \$A = 86\$ in BCD.

Similarly \$B = B1 + B0\$.

\$S\$ is \$16\$ bits
\$= S_3 S_2 S_1 S_0\$ \$= 4 \times 4\$-bit digits

So e.g. \$S_4 S_3 S_2 S_1= 0001 \space 0010 \space 0100\space 1000 \$,

Gives \$S = 0001001001001000 = 1248\$ in BCD.

The instructor wants a \$4\$ digit output from two \$2\$-digit inputs. Each digit, of course, is \$4\$ bits.

It's not entirely clear, but it looks to me like he or she expects the outputs \$S_3..S_0\$ to output the \$9\$'s complement of \$B_1 B_0 A_1 A_0\$ when \$C = 1\$ So if the two numbers are \$A = 02\$ and \$B = 12\$ then \$S = 8797\$ when \$C = 1\$.

For addition you'd expect \$S_3\$ to remain \$0\$, and \$S_2\$ to be \$0\$ or \$1\$.

If \$A = 10\$ and \$B = 1\$ then it should be \$0009\$.

If \$A = 1\$ and \$B = 10\$ then I think it should be \$9991\$.