The answer is B apparently. The way I approached this problem was to draw a free body diagram. While on the incline $$mgcos\theta$$ cancels with the normal force. So that leaves $$mgsin\theta$$ as the force that causes the acceleration while on the incline.
Work = $$Fscos\phi$$ So on the incline W = $$(mgsin\theta)(s)(1) = (mgsin\theta)(s) $$
While on the second part of the motion the only force causing the deceleration is the friction force. so $$W_2 = (-F_f)(x)(-1)$$
$$W_2 = (\mu_k)(mg)(x)$$
I just don't see where I messed up or what I am misunderstanding because apparently $W_{tot} = 0$