Fields $E$, $D$ and polarizaton $P$ I am studying Griffith's Introduction to Electrodynamics and having problems to grasp few things.
Anyway, taking into account a problem of dielectric being put between parallel plate capacitor. 
There is an E field of capacitor as an external field, and I know inside there is a polarization, dipole per volume P
With the story of induced charges and how dipoles are orienting, I believe I got the right in my mind. What I want to ask firstly, that field E and field P are in opposite direction?
I am just asking this for example of parallel place capacitor and dielectric between them?
Now, there is a final field D, which we associate with free charges. And P is associated with the bound charges.
And I should consider E fundamental and related to the total charge (free+bound).
Now: $ D=\epsilon_0E + P $
And know constantly in my head is that this should be  $-P$. Like $D+p$ should equal to $E$ and also in terms of bound and free charges.
And totally I got it in my mind that $D$ and $E$ are in same direction and $P$ is opposite.
I know question might be dump, but I am losing to understand this properly, read some previous answers here and material on Internet, but still got feeling in my head "I am no exactly sure what is happening here"...
 A: You define $E$ as an external field but $E$ in your equation is the final internal field. The final internal field is lower than the field expected if no dielectric was present. So in the standard case, $E$ is, say, positive, $P$ is positive, and $D$ is larger than $E$. 
A: You appear to be getting things turned around in your mind a bit. For typical isotropic linear dielectric materials the electric susceptibility is a positive number meaning that the P and the E fields are in the same direction: $\mathbf{P}=\chi \epsilon_0 \mathbf{E}$
For the example of a parallel plate capacitor, suppose you use a fixed amount of charge on the plates. That is a free charge, so that gives a fixed amount of D. Now, if there is no dielectric then there is a certain E field leading to a certain voltage and the capacitance is the charge divided by the voltage. Next, consider what happens if there is a dielectric. The D field is the same, but now there is a P field also. Since the E and the P field are in the same direction the E field is smaller than it was without the dielectric. This means that the voltage is smaller, so dividing the same charge by a smaller voltage gives a larger capacitance. 
So in the end D, E, and P are all in the same direction for linear isotropic dielectrics. 
A: Just considering surface charge densities on the dielectric this is a diagram of the situation described in Griffiths.
 
The important ideal is that there is an electric field in the dielectric $\vec E_{\rm local}$ which is the sum of the electric fields due to the free charges and the bound charges. $$\vec E_{\rm local} = \vec E_{\rm free} + \vec E_{\rm bound}\Rightarrow E_{\rm local} = E_{\rm free} - E_{\rm bound}$$
with the local field being less than the field due to the free charges.  
It is this local field which produces a polarization of the dielectric and the density of electric dipoles depends (linearly) on the local value of the electric field.  $\vec P = \chi \epsilon_0 \vec E_{\rm local}$ where $\chi$ is the electric susceptibility of the dielectric.
Note that since $E_{\rm free} > E_{\rm bound}$ the direction of the local field $\vec E_{\rm local} $ and hence the polarization $\vec P$ will be in the same direction as $\vec E_{\rm free}$.
So defining the displacement $$\vec D = \epsilon_0 \vec E_{\rm local} + \vec P \Rightarrow D = \epsilon_0 E_{\rm local} +  P$$  gives the relationship that you quoted in your question.  
Furthermore if $\vec P = \chi \epsilon_0 \vec E_{\rm local}$ then $D = \epsilon_0 E_{\rm local} + \chi \epsilon_0 E_{\rm local} \Rightarrow D = \epsilon_{\rm r} \epsilon _0 E_{\rm local}$ where $\epsilon_{\rm r}(= 1+\chi)$  is the relative permittivity of the dielectric.
