What is the origin of the maximum in rotational heat capacity?
Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system.  At this point (most of) the energy is being used to excite molecules from the ground state to the excited state, rather than increasing the kinetic energy (i.e. the temperature) of the molecules.

I have made video about it here.

For a more quantitative explanation, consider a two-state system with a singly degenerate ground-state and a excited state with degeneracy $g$, that is $\varepsilon$ high in energy than the ground state.  The heat capacity can be written as
$$C_V = \frac{\varepsilon^2}{RT^2}p_0p_1 $$
where $p_0$ and $p_1$ is the probability of being in the ground and excited state, respectively
$$ p_0 = \frac{1}{1+ge^{-\varepsilon/kT}}$$
$$ p_1 = \frac{ge^{-\varepsilon/kT}}{1+ge^{-\varepsilon/kT}}$$

Now some plots for $N_A \varepsilon$ = 2.5 kJ/mol and $g$ = 100



The first plot shows $p_0p_1$ (green), $C_V$ (red) and  $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis).  This shows that the general shape of $C_V$ as a function of temperature comes from $p_0p_1$



The second plot shows $p_0$ (red), $p_1$ (green), $p_0p_1$ (yellow) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis) - right $y$-axis.  This shows that the shape of  $p_0p_1$ come from the shape of  $p_0$ and $p_1$, and that the peak occurs where  the ground and exited states are equally likely ($p_0 = p_1$).  The peak in $C_V$ occurs at slightly lower temperature due to the $\frac{\varepsilon^2}{RT^2}$ term, but ultimately the peak in $C_V$ derives from the peak in $p_0p_1$.