Your question is so malformed it is hard to rectify into something meaningful. In any case, this is my attempt. It is basically but an exercise of appreciating the bra-ket notation. I doubt there is anything special about a density matrix. The direct product symbol is misplaced and possibly obstructive.
You may mean
$$
\rho=\int dx ~ |x\rangle \sigma(x) \langle x| ~.
$$
Recall what you expressed in elliptical language,
$$
P= \int dx ~ |x\rangle \frac{\partial_x}{i} \langle x| ~,
$$
as well as $\langle x|y\rangle=\delta(x-y)$, and $i\langle x|P|y\rangle=\partial_x\delta(x-y)=-\partial_y \delta(x-y)$. (Use the narrowing Gaussian limit to remind yourself of this.)
It is then evident that
$$
[P,\rho]= \int dx dy ~ \left(|y\rangle \frac{\partial_y}{i} \langle y| x\rangle \sigma(x)\langle x| -|y\rangle \sigma(y) \langle y| x\rangle \frac{\partial_x}{i} \langle x|\right )=\\
\int dx dy ~ \left(|y\rangle \frac{\partial_y \delta(x-y)}{i} \sigma(x)\langle x| -|y\rangle \sigma(y) \delta(x-y) \frac{\partial_x}{i} \langle x|\right )=\\
\int dx ~ \left(|x\rangle \frac{\partial_x}{i} ~ \sigma(x)\langle x| -|x\rangle \sigma(x) \frac{\partial_x}{i} \langle x|\right )=\int dx ~ |x\rangle \frac{ \sigma'(x)}{i}\langle x| ~.
$$
To get from the 2nd to the 3rd line, you've integrated out of $-\partial_x \delta(x-y)$ by parts. Appreciate the first derivative there (3rd line) acts on everything to its right, so its action on the bra is cancelled by the second term!
Iterating this move once again yields
$$
[P,[P,\rho]]=-\int dx ~ |x\rangle \sigma''(x)\langle x|= -2P\rho P+ P^2\rho+\rho P^2,
$$
not quite your expression, but symmetry considerations may convince you that your original desideratum expression was/is wrong. (Hint: consider the classical limit after reinstating the $\hbar$s.)
Note added in response to comment: The above linchpin expression $i\langle x|P|y\rangle=\partial_x\delta(x-y)=-\partial_y \delta(x-y)$ is a consequence of the characteristic Hermitian P action, namely $\langle x| P= -i\partial_x \langle x|$, hence $P|x\rangle=i\partial_x|x\rangle$.
Note in response to further comment by @Jyothi : This is at the heart of the bra-ket notation, which I assume you have appreciated--I can't quite give you a tutorial on it (competing with Dirac's breathtakingly superb text...). If you can start from
$P \psi(x)=-i\partial_x \psi(x)$, choosing $|\psi \rangle=|y\rangle$, you simply have the above matrix elements $ i\langle x| P| y\rangle=\partial_x\delta(x-y)=-\partial_y \delta(x-y)$, as stated.
Now multiply this on the left by $|x\rangle$ and on the right by $\langle y|$, and integrate over x and y, and then integrate by parts to get
$$
iP=\int dx dy |x\rangle \langle x|i P | y\rangle \langle y| =\int dx dy ~ |x\rangle (-\partial_y \delta (x-y) ) \langle y| \\=\int dx dy ~
|x\rangle \delta (x-y)\partial_y \langle y| = \int dx ~ |x\rangle \partial_x \langle x| ~.
$$
This is essentially a mere explanation of what the notation actually means: it transitions from the matrix bra-ket representation to functions-of-x representation, and vice versa. The reason of its popularity is that it is so compact and intuitive, once you learn the language.
