Let $m$ be a positive integer and let $\omega(m)$ be the number of distinct prime factors of $m$. The Erdos-Kac theorem, see here, is the following: For all $x \in \mathbb{R}$, $$ \lim_{n \to \infty}\left(\dfrac{1}{n} \# \left\{m \leq n \ : \ \dfrac{\omega(m) - \log\log n}{\sqrt{\log\log n}} \leq x\right\} \right) = \Phi(x) := \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{t^2}{2}} \text{dt}. $$ This means that the probability, of a positive integer $m$ having $w$ prime factors given that it was chosen uniformly at random in $[1,n]$, is asymptotically given by the Guassian: $$\mathbb{P}(\omega = w) \sim f(w; \mu, \sigma) := \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(w - \mu)^2}{2\sigma^2}},$$ where $\mu = \log\log n$ and $\sigma = \sqrt \mu$.
Time for a silly question: Since $m > 0$ by assumption, it is clear that if $w \leq 0$, then $\mathbb{P}(\omega = w) = 0$. However $f(w; \mu, \sigma) \neq 0$ for all $w \in \mathbb{R}$, and $0 \not\sim$ to anything different from $0$. What is happening?? That is the set $$ K_n(w) = \{1 \leq m \leq n \ : \ \omega(m) = w \} $$ is empty for all $n$ if $w = 0$. Hence $\frac{1}{n} \# K_n(0) \to 0$ as $n \to \infty$ but $f(0; \mu, \sigma)= \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{\mu^2}{2\sigma^2}} \neq 0$.
Not to be harsh, but the paragraph that begins "This means that..." is completely false. This simply does not follow from the Erdos-Kac Theorem. Your confusion has nothing to do with this particular result, but seems to be about weak convergence in general.
Take an easier example: let $X$ be standard normal and consider the sequence $X,X,X,\dots$. This sequence converges to a standard normal random variable, and yet the equation $$\mathbb{P}(X=x)\sim {1\over\sqrt{2\pi}} e^{-x^2/2}\tag1 $$ is false. Weak convergence makes no guarantees about $\mathbb{P}(X=x)$ for individual $x$.
As for the silly question, you have to subtract $\log\log n$ from $\omega(m)$ before you get meaningful convergence results. That is, the Erdos-Kac result makes no claim about your set $K_n(w).$