Naive Bayes Denominator clarification
The way to calculate $p(x)$ is indeed:
$$p(x) = \sum_k p(C_k) \ p(x| C_k)$$
Since in general one needs to calculate $p(C_k,x)$ (numerator) for every $k$, it's simple enough to sum all these $k$ values. It would be incorrect to use the product, indeed.

Lastly, I understand we don't actually need the denominator to find our probabilities but am asking out of curiosity.

Calculating the marginal $p(x)$ is not needed in order to find the most likely class $C_k$ because:
$$argmax_k(\{ p(C_k|x)  \}) = argmax_k(\{ p(C_k,x)  \})$$
However it's actually needed to find the posterior probability $p(C_k | x)$, that's why it's often useful to calculate the denominator $p(x)$ in order to obtain $p(C_k | x)$, especially if one wants to output the actual probabilities.