Let $K=\Bbb Q(\theta)$ be an algebraic number field with $\theta \in O_K$. Let $p$ be a rational prime. Let $$f(x)=irr_{\Bbb Q}(\theta)\in \Bbb Z[x]$$
Let $\bar{}$ denote the natural map $:\Bbb Z[x] \to \Bbb Z_p[x]$, where $\Bbb Z_p=\Bbb Z/p\Bbb Z.$ Let $$\bar{f}(x)=g_1(x)^{e_1}\cdots g_r(x)^{e_r}$$ where $g_1(x),\cdots g_r(x)$ are distinct monic irreducible polynomial in $\Bbb Z_p[x]$ and $e_1, \cdots , e_r$ are positive integers. For $i=1, \cdots,r$ let $f_i(x)$ be any monic polynomial of $\Bbb Z[x]$ such that $\bar{f_i}=g_i$. Set $$P_i=<p,f_i(\theta)>, i=1,2,\cdots,r$$.
If $ind(\theta) \not \equiv 0(\text{mod p})$ then $P_1,\cdots,P_r$ are distinct prime ideals of $O_K$ with $$<p>=P_1^{e_1}\cdots P_r^{e_r}$$ and $$N(P_i)=p^{deg f_i},i=1,2,\cdots,r$$ Here $ind(\theta)$ is the positive integer given by $D(\theta)=ind(\theta)^2d(K)$ where $D(\theta)$ and $d(K)$ are discriminant of $\theta$ and $K$ respectively.
My attempt: I have proved this one for $ind(\theta)=1$ i.e Let $K=\Bbb Q(\theta)$ be an algebraic number field of degree $n$ s.t $$O_K=\Bbb Z+\Bbb Z\theta+\cdots +\Bbb Z\theta^{n-1}$$. Let $p$ be a rational prime. Let $$f(x)=irr_{\Bbb Q}(\theta)\in \Bbb Z[x]$$
Let $\bar{}$ denote the natural map $:\Bbb Z[x] \to \Bbb Z_p[x]$, where $\Bbb Z_p=\Bbb Z/p\Bbb Z.$ Let $$\bar{f}(x)=g_1(x)^{e_1}\cdots g_r(x)^{e_r}$$ where $g_1(x),\cdots g_r(x)$ are distinct monic irreducible polynomial in $\Bbb Z_p[x]$ and $e_1, \cdots , e_r$ are positive integers. For $i=1, \cdots,r$ let $f_i(x)$ be any monic polynomial of $\Bbb Z[x]$ such that $\bar{f_i}=g_i$. Set $$P_i=<p,f_i(\theta)>, i=1,2,\cdots,r.$$
Then $P_1,\cdots,P_r$ are distinct prime ideals of $O_K$ with $$<p>=P_1^{e_1}\cdots P_r^{e_r}$$ and $$N(P_i)=p^{deg f_i},i=1,2,\cdots,r$$