Entropy and Equilibrium Constant: Chemical Thermodynamics
I agree it's not a duplicate because, while you are asking about the same homework problem discussed in the other question you linked, the issue you are having is different.  Hence, because the last answer didn't address your specific confusion, you are asking a different question about that same homework problem.

Here's where you've got things mixed up:  Whether a reaction is exothermic or endothermic depends on the inherent nature of the reactants and products, not the equilibrium constant.  I.e., $\Delta H^{o}_{rxn}$ is the enthalpy change for a complete conversion of reactants to products at standard state, irrespective of the equilibrium constant.  It's not, as I believe you were thinking, the enthalpy change to go from pure reactants to the equilibrium mixture. Hence if you could, say, change $K_{eq}$ purely by tweaking $\Delta S^{o}_{rxn}$, then $\Delta H^{o}_{rxn}$ wouldn't change just because you've changed $K_{eq}$. 

So B is correct.   If you raise T (and here we're assuming $\Delta H^{o}_{rxn}$ is independent of T), and $\Delta H^{o}_{rxn} &gt; 0$, the reason the reaction shifts to the right is this: In calculating $\Delta S^{o}_{surr}$, we have to divide $q_{rev, surr} = q_{surr}$ by T, and  T is now higher.  Hence, for the same heat flow from the surroundings (which is given by $-\Delta H^{o}_{rxn}$), the resulting unfavorable entropy change in the surroundings is less.

[I've used standard state here, which means it's carried out at a constant external pressure of 1 bar, but what I've written applies at any constant external pressure.]