Imagine a blob of liquid water. Each molecule is polar because the electrons are closer to the oxygen than the hydrogens. Without a large external electric field, the water is moving around bumping this way and that way with basically random orientations.
Now while in orbit make a very large parallel plate capacitor,charge it up and put your blob of water in between the plates. They still move around and bump into each other, but now if the hydrogen side is pointing in the direction of the electric field, it becomes harder (energy wise) to change that orientation. Over time the water can start to develop a preferred net polarization. How strong can depend on the temperature as well as the strength of the field. This is your polarization. Im your mind imagine looking at a line through the water and if the field was much stronger than the temperature induced vibrations you might see the individual charged parts of each water molecule line up like
where each +- is the two charged sides of a water molecule, so they are always always next to each other. And to someone that only cared about net charge they might look and see that it looks like
So it might look like there is only some surface charge. But that plus on the one end is bound to the negative part right by it and that negative one (on the other side of the surface) is bound to the positive part right next to it.
Now no water is pure, so you can imagine putting salt in the water and some of the NaCl crystals really do break up into Na and Cl ions (even the water itself has ions, some of the H20 molecules break up into H and OH ions) and those ions really have a net charge each and they can move around. Those are the free charges, in an external field they can move around (as charge carrier for a current, or going to the surface) and these positive and negative charges really can be far away from each other.
The bound charges aren't just affected by the parallel plate capacitor, they are affected by each other and by the free charge, but if you didn't care about the bound charges because you only care about the ions and electrons that can be added or removed from the water, then you can work with the displacement field $\vec{D}$ that ignores the bound charge. Then you get something that tracks what you care about.
An example is a high-dielectric capacitor. If all you care about is how it works as a capacitor, and you don't care about where each polarized molecule is located, then you can compute the $\vec{D}$ field inside just like for a normal capacitor.
The electric field $E$ is the field we apply, what we express with the first Maxwell equation is that its sources must come from the total density charge $\rho$. In a material, there will be some fixed charges, so the presence of $E$ will induce some dipoles, and this will make Polarization $P$ appear. Then polarization is related with the bound charge density $\rho_b$, while the rest of charges that are free to move we associate it with free density charge $\rho_f$ In vacuum, there is no bound density charge, so we have $\rho = \rho_f$.
In a presence of a material, we define displacement field $D$, or response field as
$$
D= \epsilon_0 E + P
$$
We normally consider the ideal situation for linear, homogeneous and isotropic material such that $P=\chi \epsilon_0 E$, where $\chi$ is electric susceptibility. In that way we could write $D=\epsilon_r \epsilon_0 E$, where $\epsilon_r = 1+ \chi$ is relative electric permitivity. $D$ is then taking into account the presence of free charges and bounded charges, althoug its sources are only the free charges. This can be seen easily, because the contribution of the Polarization is actually negative. It can be shown that $\nabla \cdot P =-\rho_b$, and as we know $\nabla \cdot E = \frac{\rho}{\epsilon_0}$. So if we take the divergence of the $D$ defined above and we use this results we have:
$$
\nabla \cdot D = \epsilon_0 \nabla \cdot E + \nabla \cdot P = \rho - \rho_b =\rho_f
$$
since by definition $\rho = \rho_f + \rho_b$
Best Answer
If I understand you correctly the corresponding field should be $\epsilon_0 \vec{E}$. This is one of Maxwell's equations. Let me explain a little bit about the different fields because they might be the source of confusion.
An equation you did not mention is the relation between $\epsilon$ and $\vec{P}$. We can write $\epsilon$ via the electric suceptibility: $\epsilon=1+\chi$ where $\chi$ is defined to be a proportionality factor in the following: $\vec{P}=\chi \epsilon_0 \vec{E}$. This is of course an assumption which is not necessarily always true but let us assume it is for now.
If we use that we get $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$. Taking the divergence of that one sees that $\rho_{free}=\epsilon_0\vec{\nabla}\vec{E}-\rho_{bound} $, leading to $\epsilon_0\vec{\nabla}\vec{E}=\rho_{total}$.
The equation $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$ can as far as I know be seen as the definition of $\vec{D}$ and is always true, so what I just did was some kind of backwards argument. Usually $\epsilon_0\vec{\nabla}\vec{E}=\rho_{total}$ is seen as the fundamental equation and the rest gets derived from there.
The thing which is not always true generally is $\vec{D}=\epsilon \epsilon_0 \vec{E}$ (where $\epsilon$ is a constant number). In general $\epsilon$ can be a matrix, i.e. $\vec{D}$ can have a different direction than $\vec{E}$, or it can depend on $\vec{E}$ making the relation between $\vec{D}$ and $\vec{E}$ non-linear. But nonetheless $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$ should always be true and $\epsilon_0\vec{\nabla}\vec{E}=\rho_{total}$ should also always be true as it is one of Maxwell's equations.
I hope that answers your question.