Let $P$ be a prime ideal in the ring of integers $O_K$ then $P \cap \Bbb Z=<p>$ for some prime $p \in \Bbb Z$.
So what I have to prove is that $P \cap \Bbb Z$ is prime in $\Bbb Z$.
Now $ab \in P \cap \Bbb Z$ then either $a \in P $ or $b \in P $. How do I prove that either of them are in $\Bbb Z$
What you ask at the end is not true, consider $K = \mathbb Q(i)$. Then $\mathcal O_K = \mathbb Z [i]$. Consider the ideal $P = \langle 1+i \rangle$, which is prime. Then we have that $1+i$ and $1-i$ are both in $P$ and not in $\mathbb Z$, but $(1+i)(1-i)=2$ is in $\mathbb Z.$
So you have to assume that $a,b \in \mathbb Z$.
Back to the general case. Suppose $P$ is a prime ideal in $\mathcal O_K$. Then whenever $ab\in P$ for $a,b\in \mathbb Z \subset \mathcal O_K$ we have that either $a \in P$ or $b \in P$. Thus, $P \cap \mathbb Z$ is a prime ideal. Since $\mathbb Z$ is a principal ideal domain, $P \cap \mathbb Z$ has the desired form.