# $P=\epsilon_o \chi E$ or $\epsilon_o \chi E_o$

Suppose the polarisation inside a dielectric is given by $$P$$, then is it related to the electric field as $$\vec{P}=\epsilon_o \chi \vec{E}$$ where $$E$$ is the field inside the dielectric or is is $$E$$ the original field that would have been present in that region in absence of the dielectric?

It is a matter of definition. The usual definition of $$\chi$$ implies the electric field $$\bf E$$ actually present inside the dielectric.
However, one has to notice that in the special case of a dielectric completely filling a large parallel-plates capacitor at a fixed difference of potential between the plates, the resulting field $$\bf E$$ is the same present in the vacuum when the same difference of potential exists. In this way, it is possible to control experimentally the field $$\bf E$$, which in general depends on the geometry of the sample. Moreover, since $$\chi$$ in the usual linear regime is not depending on the field, its value can be extracted in this favorable set-up, remaining the same for more complicated geometries of the external field and of the sample. Therefore, if one is referring to this special situation it may happen that the field $$\bf E$$ could be called the external field $$\bf E_0$$, although the definition contains the field $$\bf E$$.
$$E$$ is the field inside the dielectric. "Original field" is called "electric displacement" and usually denoted by $$\mathbf D$$, not $$\mathbf E$$.
Good question! $$\vec{E}$$ is the total field. i.e. the external field in the absence of the dielectric plus the field due to polarisation of charges in the dieletric.