So, it goes like this, we have Q = 0011 and Q-1 = 0 in the beginning.
At the start, we have
count = 4
(Q0 Q-1) = (1 0)
so we perform the A←(A-M) operation, according to the figure each time when we perform any operation in register A, we assume that its value is 00000000 for +M and 11111111 for −M. Performing A←(A-M) yields the first partial product
A = 11111111+00001001 = 11111001
Now, we have
count = 3
(Q0 Q−1) = (1 1)
note here, when we have (Q0 Q−1) as (1 1) or (0 0), we'll just skip and put all 0s in the partial product by shifting it by 1 bit to the left (as we do in multiplication) as it's done in the book, which is the 2nd partial product
A = 00000000+00000000 = 00000000
shifting it left by 1 bit, 00000000
Now, we have
count = 2
(Q0 Q−1) = (0 1)
so we perform the (A←A+M) operation, which gives us the 3rd partial product
A = 00000000+00000111 = 00000111
shifting it left by 2 bits, 00011100
Now, we have
count = 1
(Q0 Q−1) = (0 0)
as mentioned above, we'll just skip and put all 0s as the 4th partial product
A = 00000000+00000000 = 00000000
shifting it left by 3 bits, 00000000
(this isn't illustrated in the figure 9.14(a))
after this operation count = 0 and then we'll sum up all the partial products as
11111001 (1st partial product)
00000000 (2nd partial product)
00011100 (3rd partial product)
+ 00000000 (4th partial product)
-------------------
00010101 (final product)
As we all know (00010101)2 = (21)10. That's what done in the figure 9.14(a), but shown in a different (or COMPACT) way as
11111001
0000000
+ 000111
-------------------
00010101
Well, you can make it more compact like this:-
11111001
+ 000111
-------------------
00010101
But all of 'em will give you the binary representation of 21.