Quadratic field, $O_K/\mathfrak{p} = \mathbb{F}_p$, $O_K/pO_K$ is a finite field of order $p^2$.

It is all about the fact that $\mathcal{O}_K$ is a Dedekind domain. This is written in any book on algebraic number theory (for instance, Swinnerton-Dyer's "A Brief Guide to Algebraic Number Theory").

Applying the fact that in a Dedekind domain there is a unique factorization of ideals, we get $p\mathcal{O}=p_1^{e_1}\dots p_k^{e_k}$. So $\mathcal{O}_K/p\mathcal{O}_K=\mathcal{O}_K/p_1^{e_1}\mathcal{O}_K\dots \mathcal{O}_K/p_k^{e_k}\mathcal{O}_K$ ($e_i$ are called ramification indices). Since $\mathcal{O}_K$ is a free $\mathbb Z$-module of rank two, $\dim_{\mathbb{Z}/p\mathbb{Z}}\mathcal{O}_K/p\mathcal{O}_K=2$ so $\sum e_if_i=2$ where $f_i=\mathcal{O}_K/p_i\mathcal{O}_K$.

We can now understand the decomposition of $p\mathcal{O}_K$ looking at $\mathcal{O}_K/p\mathcal{O}_K$. Assume that $m\equiv3\bmod 4$ (the cases 1 and 2 are absolutely similar) so $\mathcal{O}_K=\mathbb{Z}[\sqrt{m}]=\mathbb{Z}[x]/(x^2-m)$ and $\mathcal{O}_K/p\mathcal{O}_K=\mathbb{Z}[x]/(x^2-m,p)=\mathbb{F}_p[x]/(x^2-m)$. Since $p$ is odd the polynomial $x^2-m$ is separable in $\mathbb{F}_p$ so it is either irreducible (in this case $\mathcal{O}_K/p\mathcal{O}_K$ is a field of order $p^2$ and $p\mathcal{O}_K$ is prime and $k=e_1=1,f_1=2$) or has two roots (in this case $\mathcal{O}_K/p\mathcal{O}_K=\mathbb{F}_p\oplus\mathbb{F}_p$ and $k=2, e_1=e_2=f_1=f_2=1$).

Finally, primes which arise in the decomposition of $p\mathcal{O}_K$ are exactly those which lie over $p$.

UPD(08.07.15) The answer of user @Servaes is more comprehensive, the fact that $\mathcal{O}_K$ is Dedekind actually doesn't required for this particular question.

As $K:=\Bbb{Q}(\sqrt{m})$ is a quadratic extension of $\Bbb{Q}$ there exists $\alpha\in K$ with $\mathcal{O}_K=\Bbb{Z}[\alpha]\cong\Bbb{Z}[X]/(f)$, where $f\in\Bbb{Q}[X]$ the (quadratic) minimal polynomial of $\alpha$. Depending on whether $m\equiv1\pmod4$ or $m\equiv2,3\pmod4$ we have $\Delta f=4m$ or $\Delta f=m$, respectively. If $p$ is a prime number that does not divide $2m$ then the discriminant of $f$ does not vanish modulo $p$, and $\Delta f$ is a square modulo $p$ if and only if $m$ is. Note that this does not depend on whether $m\equiv1\pmod4$ or $m\equiv2,3\pmod4$.

To see how this relates to the ring $\mathcal{O}_K$, note that we have isomorphisms $$\mathcal{O}_K/p\mathcal{O}_K=\Bbb{Z}[\alpha]/p\Bbb{Z}[\alpha]\cong\Bbb{Z}[X]/(f,p)\cong\Bbb{F}_p[X]/(f).$$ This shows in particular that the ideals of $\Bbb{F}_p[X]/(f)$ correspond bijectively to the ideals of $\mathcal{O}_K/p\mathcal{O}_K$, or equivalently, that ideals of $\Bbb{F}_p[X]$ containing $f$ correspond bijectively to ideals of $\mathcal{O}_K$ containing $p$.

Now if $\left(\tfrac{m}{p}\right)=1$ then $\Delta f$ is a square in $\Bbb{F}_p$, and so $f$ splits in $\Bbb{F}_p[X]$ as $f=(X-\beta_1)(X-\beta_2)$. Then $f$ is contained in the prime ideals $(X-\beta_1),(X-\beta_2)\subset\Bbb{F}_p[X]$, and no others, corresponding to two prime ideals $\mathfrak{p}_1,\mathfrak{p}_2\subset\mathcal{O}_K$ containing $p$. For each of these prime ideals we have $$\mathcal{O}_K/\mathfrak{p}_i\cong\Bbb{F}_p[X]/(X-\beta_i)\cong\Bbb{F}_p.$$

If $\left(\tfrac{m}{p}\right)=-1$ then $\Delta f$ is not a square in $\Bbb{F}_p$, and so $f$ is irreducible in $\Bbb{F}_p[X]$. Then $(f)\subset\Bbb{F}_p[X]$ is the only prime ideal containing $f$, corresponding to the only prime ideal $p\mathcal{O}_K\subset\mathcal{O}_K$ containing $p$, and $$\mathcal{O}_K/p\mathcal{O}_K\cong\Bbb{F}_p[X]/(f)\cong\Bbb{F}_{p^2}.$$

You can see Cohen's A Course in Computational Algebraic Number Theory (Theorem 4.8.13), which gives a useful algorithm for the decomposition of primes in some case, and solving your problem by using this theorem (it's an easy exercise).

Also, you can prove it straightly since it's quadratic field which is a easy case of the theorem above. There is a proof in Hecke's Lectures on the Theory of Algebraic Numbers (Theorem 89).