If \(p \le n\), then \(n!\) contains \(p\). It also means that \(2p \le 2n\), which means \((2n)!\) contains both \(p\) and \(2p\).
If \(p \gt 2n/3\), then \(3p \gt 2n\). That means \(3p\) is not in \((2n)!\). Same goes for all \(kp\) where \(k \gt 3\).
Putting the above statements together: \(2n/3 \lt p \le n\) means \(p\) and \(2p\) are in \((2n)!\) but not \(3p\) (or any higher multiple). This means \(p\) divides \((2n)!\) twice and \(n!\) once.
If \({2n \choose n} = \frac{(2n)!}{n!n!}\) and if \(p\) divides both the numerator twice and the denominator twice, then the \(p\)'s in the numerator and the denominator would cancel out. This means \(p\) can't divide \(\frac{(2n)!}{n!n!}\).