Carry Generator Gi=Ai.Bi and Carry Propogater Pi=Ai xor Bi
lets take an example with 4 bits to make it clear
G0=A0.B0 and P0= A0 xor B0
G1=A1.B1 and P1= A1 xor B1
G2=A2.B2 and P2= A2 xor B2
G3=A3.B3 and P3= A3 xor B3
we know that C0 is initial carry
C1=C0.P0 + G0 ------(a)
C2=C1.P1 + G1 -------(b)
C3=C2.P2 + G2 -------(c)
C4=C3.P3 + G3 ---------(d)
substitute (a) in (b)
C2=C0.P0.P1+G0.P1+G1 ----- (e)
and (e) in (c), and so on till you get the C4 i.e.,
C4=C0.P0.P1.P2.P3+G0.P1.P2.P3 + G1.P2.P3 + G2.P3 + G3
if you clearly observe C4= (4 AND gates) + (3 AND gates) + (2 AND gates) + ( 1 AND gate) + ( 0 GATES)
it is 4+3+2+1=10. In general for n-bit carry look-ahead adder, n(n+1)/2 AND gates are needed.
We can easily say from the C4 that only n no.of OR gates are required.
so Answer is 7*8/2=28 AND gates and 7 OR gates.