Answer
b) and d) are correct.
Work Step by Step
According to the principle of energy conservation, $$E_f=E_0$$ $$\frac{1}{2}mv_f^2+mgh_f=\frac{1}{2}mv_0^2+mgh_0$$ $$\frac{1}{2}m(v_f^2-v_0^2)=mg(h_0-h_f)$$
When the object moves uphill, we have $(h_0-h_f)\lt0$. For the equation to work, it follows that $(v_f^2-v_0^2)\lt0$, or the object has a decreasing speed. (b) is correct.
When the object moves downhill, we have $(h_0-h_f)\gt0$. For the equation to work, it follows that $(v_f^2-v_0^2)\gt0$, or the object has a increasing speed. (d) is correct.
