Yes and no. In this particular doc the change you are seeing reflects the change of uncertainty of the estimation, but not the posterior.
For a similar model you can estimated the posterior uncertainty by increasing PPC, because for Bernoulli the variance (or uncertainty) is linked to the mean, but in general it is not the case.
Think of a Gaussian, where we have better intuition:
Say you have a data generation process y \sim \text{Normal}(\mu, \sigma^2), with \mu and \sigma being the model parameters. When both parameters are fixed, the uncertainty of y is fully explain by the property of Gaussian, and its uncertainty is quantify by a function of \sigma (e.g., if you quantify the uncertainty as variance to the mean then it is just \sigma^2).
Now, say you have \sigma fixed and \mu from some distribution (in PPC, it would means \mu \sim \pi_{posterior}(...)), then the distribution for y becomes a result of a convolution (see Colah’s blog on this where in the beginning he gives an very intuitive representation). Which to answer your initial question, the posterior uncertainty is in \pi_{posterior}(...), where as PPC uncertainty is the convolution of \pi_{posterior}(...) and a Gaussian with the fixed \sigma
All these is to show you that, increasing PPC sample does not necessary give you the ability to quantify the uncertainty of posterior for \mu, but it can let you better estimate of the uncertainty of y.