$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
Let's suppose that the reaction is taking place at constant pressure (for constant volume, see the last paragraph). In that case, the most likely state functions to be represented on the vertical axis are H and G. Then:
$$\pd{G}{T}{p} = -S$$
$$\pd{H}{T}{p} = C_p$$
From this we can see that the vertical coordinate can't be G. That's because your diagram has the dependent variable increase with temperature. But, since $S>0$ (always), $\pd{G}{T}{p} <0$ (always).
Now, what about the sign of $C_p$? Well, $C_p=\frac{\text{đ}q_p}{dT}$. If you flow heat into a system ($\text{đ}q > 0$) without making any other changes, the temperature will always increase. And vice versa if heat flows out of a system. Thus the signs of $\text{đ}q_p$ and $dT$ will always be the same $ \implies C_p > 0 \implies \pd{H}{T}{p} >0$.
Thus, by elimination (and based on our starting assumption), we conclude the vertical coordinate is H.
In that case, we would expect the temperature-dependence of the enthalpies of the reactants, activation complex, and products to be relatively small—the main effect of increasing temperature would be to increase the rates of the forward and reverse reactions, by increasing the proportion of reactants and products able to pass over the activation barrier, as well as the collision frequency.
However, to the extent that you do increase the temperature enough to see a temperature-dependence in enthalpies, it would increase the enthalpies of everything—the reactants, products, and activation complex. Thus, qualitatively, the blue curve would be correct.
I say "qualitatively" because the heat capacities of the reactants, products, and activation complex will all be different, and thus the extent to which their respective enthalpies are raised by an increase in temperature will likewise be different.
Note that the same argument would apply if we were at constant volume instead of constant pressure, in which case the most likely state functions to use for the vertical axis would be either U (the internal energy) or A (the Helmholtz free energy). And here the only one of these two whose value could increase with temperature (at constant volume) would be U.
$$\pd{A}{T}{V} = -S < 0$$
$$\pd{U}{T}{V} = C_V > 0$$
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