Assuming that E is planar, then it can be antiaromatic. Since the boron has an incomplete octet and the neighboring nitrogen has a pair of electrons to share, you can draw a double bond between these atoms (though formal charges will exist on the boron and nitrogen, that won't matter). Now the structure will have 4 double bonds or 8-pi electrons, so it will be antiaromatic.
However, just as cyclooctatetraene would be antiaromatic, if planar, it is actually non-aromatic. Because compounds that are antiaromatic are higher in energy, if a structure can relieve unfavorable interactions by conformational changes, they probably do so.
Compound B is non-aromatic as the nitrogen is locked into an sp3 hybridization and thus cannot participate in any resonance structures.