Question: The $10$ generators of the Poincare group $P(1;3)$ are $M^{\mu\nu}$ and $P^\mu$. These generators can be determined explicitly in the matrix form. However, I have found that $M^{\mu\nu}$ and $P^\mu$ are often written in terms of position $x^\mu$ and momentum $p^\mu$ as

$$ M^{\mu\nu} = x^\mu p^\nu - x^\nu p^\mu $$
and
$$ P^\mu = p^\mu$$

How do we get these relations?

Answer: 
One obtains those expressions by considering a particular action of the Poincare group on fields.

Consider, for example, a single real scalar field $\phi:\mathbb R^{3,1}\to\mathbb R$. Let $\mathcal F$ denote the space of such fields. Define an action $\rho\_\mathcal F$ of $P(3,1)$ acting on $\mathcal F$ as follows
\begin{align}
 \rho\_\mathcal F(\Lambda,a)(\phi)(x) = \phi(\Lambda^{-1} (x-a))
\end{align}
Sometimes people will write this as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity. Now let $G$ denote a generator of the Lie algebra of the Poincare group (namely an element of a chosen basis for this Lie algebra). We can use this generator to define a corresponding infinitesimal generator for group action $\rho\_\mathcal F$ as follows:
\begin{align}
 G\_\mathcal F(\phi) = i\frac{\partial}{\partial\epsilon}\rho\_\mathcal F(e^{-i\epsilon G})(\phi)\bigg|\_{\epsilon = 0}
\end{align}

**Example - translations.** Consider the translation generators $P^\mu$ which have the property
\begin{align}
 e^{-ia\_\mu P^\mu}x = x+a
\end{align}
The generator of $\rho\_\mathcal F$ corresponding to $P^0$, for instance, is
\begin{align}
 (P^0)\_\mathcal F(\phi)(x) 
&= i\frac{\partial}{\partial\epsilon}\rho\_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|\_{\epsilon = 0} \\
&= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e\_0)\bigg|\_{\epsilon = 0} \\
&= i\partial^0\phi(x)
\end{align}
where $e\_0 = (1,0,0,0)$, and similarly for the other $P^\mu$, which gives
\begin{align}
 (P^\mu)\_\mathcal F = i\partial^\mu.
\end{align}

**Example - Lorentz boosts.**

If you use this same procedure for Lorentz boost generators, you will find that
\begin{align}
 (M^{\mu\nu})\_\mathcal F = i(x^\mu\partial^\nu - x^\nu\partial^\mu) = x^\mu p^\nu - x^\nu p^\mu
\end{align}

**Disclaimer about signs etc.** There are a lot of conventional factors of $i$ and negative signs floating around which I wasn't super careful to keep track of, if you notice an error in this regard, please let me know and I'll fix it.
The generators of isometry, also generators of the Poincare group, are the Killing vectors, hence we need the Killing vectors of Minkowski spacetime, $ds^2 = -dt^2 + dx^2 + dy^2+ dz^2$. The defining equation of the Killing vectors in terms of their components ($\xi = \xi^\mu \partial\_\mu$ is a Killing vector with components $\xi^\mu$) is $$\partial\_{(\mu} \xi\_{\nu)}=0,$$ where the parentheses denote symmetrization over the enclosed indices. We differentiate once, then cyclically permute the indices, $$\begin{split}\partial\_c\partial\_a \xi\_b + \partial\_b\partial\_c \xi\_a &=0,\\
\partial\_a\partial\_b\xi\_c + \partial\_c\partial\_a\xi\_b&= 0,\\
\partial\_b\partial\_c\xi\_a + \partial\_a\partial\_b\xi\_c &=0,\end{split}$$ which is a linear system with unknowns $\partial\_a\partial\_b\xi\_c$ and its permutations. The only solution of the system is the trivial $\partial\_a\partial\_b\xi\_c=0$, from which we obtain $\xi\_a = a\_{ab} x^b + b\_a$ with $a\_{ab}$ and $b\_a$ constants. From the defining equation we obtain $a\_{ab} = -a\_{ba}$ and, since the Killing vectors are unique up to multiplication with a constant and translation, we obtain $$ \begin{aligned}
 \partial\_{t} && \partial\_{x} && \partial\_{y} && \partial\_{z},\\
 -x\partial\_{t} - t\partial\_{x}, && -y\partial\_{t} -
 t \partial\_{y} && - z\partial\_{y} - t\partial\_{z},\\
 x\partial\_{y} - y\partial\_{x} && y\partial\_{z} - z\partial\_{x} &&
 z\partial\_{x} - x \partial\_{z}.&&
 \end{aligned}$$
On the first line are the translation generators, on the second the boost generators and on the third the rotation generators.