From a physics perspective, for a capacitor we have
$$Q = Cv$$
Where \$Q\$ is the amount of charge separated (\$Q\$ charge on one plate, \$-Q\$ charge on the other), \$C\$ is the capacitance and \$v\$ is the voltage across the capacitor.
Due to conservation of electric charge, if \$Q\$ is changing, there must be a current \$i\$ into one plate and out of the other thus
$$i = \frac{dQ}{dt} = C\frac{dv}{dt}$$
Note that when the voltage across a capacitor is constant, i.e., \$\frac{dv}{dt} = 0\$, the capacitor current is zero.
Also note that when the capacitor current is constant, then the rate of change of capacitor voltage is constant.
Now, with that review in mind, consider that the equations above do not imply that a capacitor "blocks DC".
Rather, they imply that, for a DC (constant) voltage, the capacitor current is zero.
And, for a DC (constant) current, the capacitor voltage steadily changes.
But, if the voltage is changing, there is a changing electric field and thus, a changing electric flux in the dielectric of the capacitor.
And, according to Maxwell's equations, a changing electric flux is a type of electric current and produces a magnetic field just as a conduction current does:
$$\nabla \times \vec H = \vec J + \frac{\partial \vec D}{\partial t} $$
where the first term on the right hand side is the conduction current density and the second term is the electric flux current density (displacement current density).
So, it is true that a time varying capacitor voltage is associated with a displacement current in the capacitor dielectric and, in fact, this must equal the conduction current into and out of the capacitor.
But, as I've pointed out, for a fixed capacitor voltage, there is no conduction or displacement current.
