Polarization P is caused by the presence of E and a dielectric material. It does not "add" to E, instead it adds to flux D. It does not relate to free charges. And the relationship is
$$ D = \epsilon_0 E + P. $$
Polarization reflects what happens to the bound charge-pairs in said dielectric (i.e. the amount the charge-pair separate with applied electric field). By virtue of the separation of the charge-pairs, there is a localized E field created, which distort E locally but since net charge of each pair is still 0, it does not add E globally, And it does not affect D outside the material.
One encounters P in the simple and most common case of linear dielectrics, as contributing to the relative permittivity. That is,
$$ D = \epsilon_0 E + \epsilon_0(\chi)E. $$
where $\chi$ is called electric susceptibility.
Or to be more accurate, we create a mathematical constant, relative permittivity, i.e.
$$ \epsilon_r = \frac{P}{\epsilon_0 E} - 1, $$
so we can more easily relate P, E and D, vis a vis
$$ D = \epsilon_r \epsilon_0 E. $$
With regards to boundary condition, other than changing relative permittivity on either side of the boundary, there is no other consideration specific to P. All boundary condition of E and D still hold, regardless of P (or $\epsilon_r$).