You don't say which ideal in $O$ you are trying to decompose. I will guess you mean to decompose the ideal $pO$. The number of primary ideals in a *minimal* primary ideal decomposition of $pO$ in $O$ *need not* be the same as the number of prime ideal factors of $pO_K$ in $O_K$.

Example: Pick your favorite prime number $p$ and let $O$ be the ring ${\mathbf Z} + pO_K$. Set ${\mathfrak b} = pO$ and ${\mathfrak P} = pO_K$. Obviously ${\mathfrak b}$ is an ideal in $O$, but note that ${\mathfrak P}$ is also an ideal in $O$ (it's an ideal in the bigger ring $O_K$ which happens to lie in the ring $O$, so it's also an ideal in $O$) and ${\mathfrak P}$ is a prime ideal in $O$ (not necessarily in $O_K$!) since $[O:{\mathfrak P}] = [O:pO_K] = p$ is a prime. Since, set-theoretically, ${\mathfrak b}$ contains ${\mathfrak P}^2 = p^2O_K$ and is contained in ${\mathfrak P}$, ${\mathfrak b}$ is a primary ideal, and thus the minimal primary ideal decomposition of ${\mathfrak b} = pO$ in $O$ is ${\mathfrak b}$ itself. I have put no conditions on $p$ at all here (though we chose $O$ depending on $p$) so for instance you could now let $p$ be a prime number which splits completely in $O_K$ (there are infinitely many of those), and in that case your $m$ is $n$ and that is not the number of ideals in the minimal primary decomposition of ${\mathfrak b} = pO$ in $O$ (well, assuming $K \not= {\mathbf Q}$, so $n > 1$).

The subtlety here is related to the conductor of your ring $O$. The answer to your question would be $m$ for prime numbers that are relatively prime to the conductor of the ring $O$ you use. In the example above, the conductor of $O$ is $pO_K$, which contains $p$ and thus one can anticipate complications.

Perhaps it would help if you explained why you are asking this question.