Effect of temperature on equilibrium constant in terms of entropy change
The key point is that the entropy of the surroundings changes, not that of the system (at least for a small change in $T$), so only [B] and [D] are relevant, as you correctly determined.
You can write
$$\log(K)=-\frac{\Delta H}{RT}+\frac{\Delta S}R$$
This of course is one form of van 't Hoff's equation. It tells you, assuming $\Delta H$ and $\Delta S$ are constant, that the temperature dependence is due to $\Delta H-$ the heat transferred to/from the surroundings (entropy increase/loss of the surroundings):
$$\log(K)\propto-\frac{\Delta H}{RT}=\frac{\Delta S_\text{surr}}R$$
Now increasing $T$ makes the change in the entropy of the surroundings smaller.
When $\Delta H\gt0$ (endothermic), ie $S_\text{surr}\lt0$, this makes the reaction more likely to happen. When $\Delta H\lt0$ (exothermic), ie $S_\text{surr}\gt0$, this makes the reaction less likely to happen.
So both [B] and [D] are correct.