Why Does Activation Energy not change with Temperature change As The Graph Shows; $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants.
So let This be the graph at Temperature $T_1$, Now say We Increase The Temperature to say $T_2$ and so the energy of reactants must increase (as kinetic energy of molecules increases). So at $T_2$, reactants would start with more energy. So Shouldn't the Activation Energy decrease as now less energy is needed to reach Transition State.
Or I can say that since $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants, So as Energy of Reactants has changed so should Activation Energy.
I already saw answers on a similar question but couldn't figure out what's wrong with my approach.
You have conceptual misunderstanding. In the activation energy context, energy of reactants, products and transition state are all potential energies, independent on the kinetic energy of molecules.
The kinetic energy of molecules is used to reach the peak of the potential energy, similarly as e.g. a acrobatic aircraft uses its kinetic energy to climb to the apex of an acrobatic loop.
Higher temperature means higher kinetic energy, but it does not mean the potential energy barrier is lower. It means there is higher probability to overcome it, leading to a faster reaction.
There is a famour Arrhenius' equation for the reaction rate:
$$k = A \exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$$
where $A$ is the frequency factor, describing the rate of molecular collisions, potentially able to take part of the reaction and
$\exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$ is the term from the Boltzmann distribution,
telling us the probability a molecule at temperature $T$ has sufficient energy, if the activation energy is $E_\mathrm{A}$, and $k$ is the Boltzmann constant.