CDF and PDF in the case of normal distribution I was reviewing some really basic probability and came across a statement that confused me.

We write X ∼ N(μ, σ^2) to denote that p(X = x) = N(x|μ, σ^2). If X ∼ N(0, 1), we
say X follows a standard normal distribution.
We will often talk about the precision of a Gaussian, by which we mean
the inverse variance: λ = 1/σ^2. A high precision means a narrow
distribution (low variance) centered on μ. Note that, since this is a
pdf, we can have p(x) > 1. To see this, consider evaluating the
density at its center, x = μ. We have N(μ|μ, σ^2) = (σ√2π)−1e^0, so if σ
< 1/√2π, we have p(x) > 1.

Here, I can get my head around everything except that I am not sure how to interpret the case where p(x) > 1.
If I'm reading the statements correctly, as variance gets smaller, the probability of mean should get higher,which makes sense, but I do not see how the probability exceeds one.

Because $p(x)$ doesn't indicate a probability but  "probability density". In fact for a continuous rv you have a pdf and not a pmf .
The probability is defined only in a range $(a;b)$ and it is the integral $\int_a^b p(x)dx$
This is an example of Gaussian density with mean 0 and $\sigma=0.1$

