Why is there no Bessel's correction equivalent for $\sigma^2 = \sum [(x - \mu)^2 P(x)]$? Why is sample variance and discrete variable variance different in this sense? Is it because Bessel's correction exists for a better reflection of the population variance whereas $\sigma ^2$ already represents the population of the discrete variable?

For a discrete population $$\sigma^2 = \mathbb E\left[(X - \mu)^2\right]  = \sum_x (x - \mu)^2 P(X=x)$$ is the definition of variance.  No further adjustment is necessary
Bessel's correction can apply when you use the sample mean in the calculation of the sample variance because $$\mathbb E\left[ \frac1n\sum_i (x_i -\bar x)^2\right] = \frac{n-1}{n}\sigma^2$$ and you may prefer an unbiased estimator of the population variance
