T(n)=T(n-1)+T(n-2)+1
For simplicity, we can write like this...
T(n)=2T(n-1)+1
a=2, b=1, k=0
According to Muster's theorem
T(n)= O($n^{k}a^{\frac{n}{b}}$) = O($2^{n}$)
For more about muster theorem- https://www.geeksforgeeks.org/master-theorem-subtract-conquer-recurrences/
The above problem can also be solved using the dynamic approach, in which we don't have to compute T(n-1) two times in the given recurrence relation. We can compute for the 1st time and store it for the next. In this way time complexity would be...
T(n) = T(n-1)+1
a=1, b=1, k=0
According to Muster's theorem
T(n)= O($n^{k+1}$) = O($n^{1}$)
So C is the right answer.