How can I minimize floating point error when multiplying normal distribution PDFs?
While there is an entire subject that studies how to prevent or reduce floating point errors from piling up, it happens we have a shortcut here.

Let $\mathcal N(u, \sigma)=\frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$ , the normal distribution with mean $u$ and standard derivation $\sigma$.
Suppose we are given $n$ Gaussian normal distribution $\mathcal N_i(u_i, \sigma_i)$ . What is the product of $n$ of them? It turns out,

$$\prod_{i=1}^n \mathcal N(\mu_i,\sigma_i)= s\mathcal N (\mu,\sigma)$$ 
where
$$
\frac1{\sigma^{2}} = \sum_{i=1}^n \frac1{\sigma_i^{2}}\,, \quad
\frac\mu{\sigma^2} = \sum_{i=1}^n \frac{\mu_i}{\sigma_i^{2}}
$$
and
$$
s =  (2\pi)^{\frac{1-n}2} \frac{\sigma}{\prod_{i=1}^n \sigma_i}\exp\left(\frac12 \left(\frac{\mu^2}{\sigma^2} - \sum_{i=1}^n \frac{\mu_i^2}{\sigma_i^2}\right) \right)
$$

If you use the above formula to compute, floating point errors should not be a concern any more. 


Please note the reference you used is misleading in the sense the product function is not a probability distribution function.



Exercise. Prove the above formula for the product of $n$ normal distribution.