Relationship between a temperature increase of an exothermic reaction, Keq, and kf (forward rate constant)
You know that both the forwards $k_f$ and reverse $k_r$ rate constants will increase with temperature, simply because there is more energy present and so it takes less energy to surmount the activation barrier, both from the product and reactant sides. 

The question is then which is changed more by changing the temperature. The equilibrium constant is as you write $K_Q=k_f/k_r$ and now using the Arrhenius equation we get 
$$K_Q=\frac{A_f\exp(-\Delta E/(RT))}{A_r\exp(-(\Delta E+E_0)/(RT))}$$

where the A's are pre-exponential factors (and are constant vs temperature) and $\Delta E$ the activation barrier for the forwards reaction and $E_0$ the extra energy for the back reaction and  is positive since the reaction is exothermic. (If you sketch the barriers hopefully this should be clear).

Simplifying gives 
$$K_Q=\frac{A_f}{A_r}\exp(E_0/(RT))$$

As T  is increased $K_Q$ becomes smaller. Thus the answer is (D).