Why is the polarity of the B-N bonds in borazine the reason for its reduced delocalisation of π electrons?
Let us understand and try to answer your question with a few simple points.
First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons.
According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for allowing the electrons of the side atom to delocalize into it needs to be satisfied. Along with this, the lone pair which needs to be delocalized should be in kept in a pure p-orbital. Boron and Nitrogen both facilitate these requirements and thus the lone pair delocalization takes place.
Thus in a way, we could say that the aromaticity in the borazine moeity is due the pi-backdonation of unpaired electrons.

Understanding that the main reason is sideways pi-backdonation for the cyclic delocalization of electrons, one also needs to keep in mind that Boron and Nitrogen have a significant difference in their electronegativities. The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene. You could similarly compare h-BN and grpahite; in the former case the delocalisation is less pronounced due to difference in electronegativity than in the later case where a very well pronounced delocalisation is observed.