@avu017 this is basically two separate problems, solve for a and b:
2 sin a = 2 <==> sin a = 1 <==> a = pi/2 + (2k*pi) where k is an integer (to solve quickly, look at the unit circle and note that sin(pi/2) = 1. You can add any multiple of 2pi and the sine of that angle will still be 1).
2 cos (pi/2 - b) = 2 <==> cos (pi/2 - b) = 1 <==> b = pi/2 + (2m*pi) where m is an integer
Adding the two solutions, we have a+b = pi + (2(k+m)*pi) where k and m are integers, and only D) is a possible value.