Why is a bandpass filtered analog PM signal still PM and not AM
Conventional AM and narrowband PM signals look quite similar:
$$x_{AM}(t) = A \cos(\omega_c t) + m(t) \cos(\omega_c t) $$
$$x_{NBPM}(t) = A \cos(\omega_c t) - A k_p m(t) \sin(\omega_c t) $$
And based on this, their spectrum also look quite similar. However, there's an important distinction between them:
AM modulation changes (modulates) the amplitude of the carrier wave, but keeps carrier phase unaltered. Whereas, PM modulation, NW or WB, keeps the amplitude of the carrier constant, but alters (or modulates) its phase according to the message signal.
Ther reason for this different behaviour is about the way the sidelobes (spurs) are added to the main carrier. In AM, the sidelobes are added in-phase with the carrier (using two cosines for that), while in NBPM they are added in quadrature (sine added to cosine). This quadrature additon, leaves the amplitude (almost) unaltered but changes the phase-angle of the carrier.
So if you treat NBPM as an AM and try to demodulate it that way, you will get constant output (or nothing after the DC blocking), as the amplitude of the NBPM carrier is not changing...