When A=1, then outp put Q will be selected by MUX and feedback to D flipflop which gives output Q again. So at A=1 it holds its state.
But, when A=0, then \overline Q will be selected by MUX and feedback to D flipflop and output will be inverted. And after every clock pulse output will be toggled due to section of \overline Q.
So option (d) is correct option.