Bounded charges density different from zero From wikipedia, the electric displacement field is $\vec D=\epsilon_0 \vec E+ \vec P$ and it satisfies $\nabla \cdot D=\rho-\rho_b$ where $\rho_b$ is "the density of all those charges that are part of a dipole, each of which is neutral".
If these equations works at macroscopic level (by which i mean that every quantity is the result of an average over a small volume containing a huge number of particles) then why $\rho_b$ is not zero?
If any particle is neutral then every small volume contains zero net charge and so $\rho_b \triangleq dQ/dV=0$
 A: As you stated:

$\rho_b$ is "the density of all those charges that are part of a dipole, each of which is neutral".

This means that the dipole as a whole is neutral.
But there can still be a charge distribution with non-zero values internally for the dipole. For example, two charges of equal magnitude but opposite value (e.g., $q$ and $-q$) separated by a distance $d$.

Why ρb is not zero? If any dipole is neutral then at macroscopic level it should always be zero

The dipole is not zero (not even "at macroscopic level"). The total charge of the dipole is zero, but the dipole is not "zero"--it is a dipole.

Update, based on chat:
It is true that you could can often choose some volume where the total charge is zero, but that is not the only thing that matters. For example, suppose you have a whole bunch of dipoles lined up in a row (which you actually can/do have in macroscopic matter). Inside the material there are a lot of different volumes that you could choose where the charge inside the volume is zero.
But, nevertheless, at the surfaces, there must be bound charge (think about a whole line of dipoles all in a row facing the same direction, at the very ends of the line there is uncompensated charge). (In a similar polarized volume this would mean that the bound charge is vanishing except for at the surface.)
