How does a Miller cap physically create a pole in circuits?
However, physically, there is no path between the input (gate) and the ground 


Firstly, you do not need a capacitor connected to ground to create a pole in a system. Just the presence of the capacitor wherever it is connected is sufficient. A simple example is this high pass filter



This circuit has a transfer function given by:

$$H(s) = \frac{sCR}{1+sCR}$$
So we have a pole even though the capacitor is not connected to ground.

So, how are poles created?
One way to think about it is that the impedance of a capacitor (or an inductor) is given by \$1/sC\$ (\$sL\$), so whenever we have a these elements we will get a pole. Because it will result in the transfer function where the coefficients of terms with \$s\$ will be non zero.
Thus poles are created when we have energy storing elements present in a circuit. A capacitor will store energy in the form of voltage and the inductor in the form of current. The number of poles will be given by the number of indendent energy storing elements. For instance a capacitor in parallel with R1 in the above circuit will not create additional pole but if we cascade this system with another R and C, we will get a second order system.

Now, it should be clear that the miller capacitor will add to a pole. It's also easy to see why the capacitor is multiplied by gain. Since the capacitor is connected across an inverting gain amplifier, if the input of the amplifier rises by \$\Delta V\$ then the output would go down by \$A\Delta V\$. Thus the charge stored is \$\Delta Q = C(1+A)\Delta V\$. Thus, \$C_{eq} = \frac{\Delta Q}{\Delta V} = (1+A)C\$