I know my answer is not interactive much but this explanation can help u understand the answer well
First the time delay to calculate the value of P's and G's = 1 time delay(all P's and G's would be calculated parallely)
as Pi = Ai ex-or Bi and Gi = Ai.Bi
now sum at last would be calculated in 1 time delay as sum = Pi ex-or Ci ( all such ex-or's for S0, S1, .. would be done parallely)
So total time = 1(P's and G's) + carry generation time + 1(Pi ex-or Ci)
now carry generation time would be max for maximum carry as for 4 bits we need C4 and that will take max time
so time to generate carry = time to generate Cn for n bits
lets say n = 6
then C6 = g5+ P5G4 + P5P4G3 + P5P4P3G2 + P5P4P3P2G1 + P5P4P3P2P1G0 + P5P4P3P2P1P0C0
calculation will be done parallely so biggest term here P5P4P3P2P1P0C0 would take 3 time delay becoz 3 levels of calculation
2nd term = 1 time delay = ceil(log2(2))
3rd term = 2 time delay = ceil(log2(3))
4th term = 2 time delay = ceil(log2(4))
5th term = 3 time delay = ceil(log2(5))
6th term = 3 time delay = ceil(log2(6))
7 th term = 3 time delay = ceil(log2(7))
so max time delay in multiplication and some addition as well = 3 = ceil.log2(7) = ceil.log2(n+1)
Some addition would be performed parallely as here in this example addition of any 2 terms of first 3 can be done in parallely with in 3 time delays but what about all terms with 3 time delay they would take further time to add here max 4 terms can be there with 3 time delay that would take log2(4)- no of terms in max time delay here 3 max time
so how many max terms with ceil(log2(n+1)) = 2^(ceil(log2(n+1))-1)
now addition will be performed on these max terms that would take log2(maxterms) = ceil(log2(n+1) -1 time
so carry generation time = ceil(log2(n+1))+ ceil(log2(n+1))-1 = 2 ceil(log2(n+1)) -1
so total time = 2 ceil(log2(n+1))-1 +2
= 2ceil(log2(n+1)) +1