In my understanding of a Naive Bayes Classifier, one takes the argmax of the probabilities that example $x$ belong to class $c_i$, that is $$\text{argmax}_{c_i\in C}P(C=c_i|X=x)$$
I understand that $P(C|X)$ can be calculated/estimated using Bayes theorem: $$P(C|X) = \frac{P(C)P(X|C)}{P(X)}$$
The wikipedia article on this subject suggests that the probability $P(X|C)$ can be replaced by the evaluation of a probability density function $p$ (in their example, of a normal variable with $\mu$ as the mean of the data set and $\sigma$ the standard deviation) so the evaluated "probability" (is it really still a probability?) becomes: $$p(C|X) = \frac{P(C)p(X|C)}{P(X)}$$
I don't understand how one can replace a probability with a pdf. In my understanding of what a pdf is, it's evaluation at a given point doesn't feel right.
Why is this still valid?
I do not see any contradiction. Let me try to answer without resorting to formal probability concepts.
In Naive Bayes classifier, the conditional random variable $X|\{C=c\}$ is assumed to be multivariate gaussian with a diagonal covariance matrix (a.k.a independence among the coordinates of vector $X$ is assumed, though this is not important for the discussion here). The random variable $C$ is assumed discrete.
Your confusion is applying Bayes rule when both $X$ and $C$ are discrete. In this case, $$ \mathbb{P}(C = c| X = x) = \frac{\mathbb{P}(X = x| C = c)\mathbb{P}(C = c)}{\mathbb{P}(X = x)} $$
But in our case, we use the version of Bayes rule when $C$ is discrete and $X$ is continuous. This is defined as (See link): $$ \mathbb{P}(C = c| X = x) = \frac{f_{X}(X = x| C = c)\mathbb{P}(C = c)}{f_{X}(X = x)} $$ where $f_{X}(x)$ is the probability density function (pdf) of $X$.