This is a nice question and requires redrawing the circuit..However before redrawing the circuit we see that the enable pin being used is an active low pin..So when c = 0 which is fed to enable pin..
So we have minterms 0 , 2 , 4 and 6 only for which decoder will select and give result..
Now coming to redrawing circuit , we see we have active low output as well.So after readjustment of bubbles in the circuit after decoder we will have active high output of decoder followed by AND gate instead of OR gate and OR gate instead of AND gate followed by OR gate instead of NAND gate finally..
So the AND gate which is ANDing minterms {0,2} will evalaute to 0 as this is a property of decoder that exactly one output pin will be high at a time..
Hence we are left with the output of OR gate which comprises of minterms {4,6} ..So the final result will be : Σ(4,6)
Hence C) should be the correct answer..
