Why does the electric polarisation (polarisation density) have this direction? It can be inferred that the amount polarisation $ P $ is dependent upon and proportional to the magnitude of the electric field $ E $ applied:
\begin{equation}
 P \propto E.
\end{equation}
This makes sense from what we know about the nature of electric polarisation and is also supported by the well-known relationship that
\begin{equation}
 \mathbf{P} = \epsilon_0\chi_e \mathbf{E}.\ \ \ \ \ \ \ (1)
\end{equation}
However, the electric field will push positive charges in the direction of the field and negative charges against the field direction. 

Convention dictates that vectors regarding electric charge must point from the positive charge to the negative, implying that
\begin{equation}
 \hat{\mathbf{P}} \propto - \hat{\mathbf{E}},
\end{equation}
which we know to not be true due to equation 1 (both $\epsilon_0$ and $\chi_e$ are always greater than 0.)
What is the justification for the vector direction that allows $\hat{\mathbf{P}}=\hat{\mathbf{E}}$ by equation 1 when the arrangement of charge suggests the opposite?
 A: The reason is simple: the convention is that the polarization vector goes from negative to positive, not the other way around.
A: Imagine two opposite charges very very close together
In the tiny region between the charges the field goes from the positive charge to the negative charge. However if you come towards the positive charge from the opposite direction as the negative charge the field points in the opposite direction.
And if you come towards the negative charge from the opposite direction as the positive charge the field points in the opposite direction.
Why? Because both charges exert forces, but the $q/r^2$ gets weaker the farther away you are so you still point away from the positive charge and towards the negative charge.
Now if you imagine the distance between the charges as getting smaller eventually it looks like the field just comes into the dipole over where the negative charge was and jumps out the other side where the positive charge was.
That vector is the polarization vector. Check out the gif at wikipedia https://en.m.wikipedia.org/wiki/Electric_dipole_moment#/media/File:VFPt_dipole_animation_electric.gif
Now a real electric dipole has that field pointing the opposite direction in between you can approximate it with an ideal dipole which doesn't have that space between the charges.
If it helps, you can think that the vector pointing from negative to positive is your reminder that you are approximating it as if the charges were on top of each other (closer than they are really) and you changed how strong the charges were. Not because it was accurate, but because that was pretty close to the correct field when you are not very close.
You always want to know when you are approximating, so the polarization vector pointing that way can tell you that you are approximating as if there were ideal dipoles there.
A: The microscopic field is directed from positive to negative. Ie, opposite to the applies field. But those field lines are closed loops. Their net field is zero.
The only field that contributes to net field is the field that enters the dipole and that leaves the dipole. That field is in the direction of the applies field.

If we consider entire dipole as one gaussian surface, the field enters the negative end and leaves the positive end. So the net field direction is from negative to positive, though the microscopic field is directed from positive to negative.
That is why Polarization (Net induced field) is directed opposite to microscopic field direction.
