From De Broglie’s relations and the energy-momentum dispersion relation one finds: $$v_p=\frac{w}{k}=\frac{E}{p}=\frac{\gamma mc^2}{\gamma mv}=\frac{c^2}{v} \tag{1}$$ Where, $v_p$ is the phase speed of the wave, $v$ is particle’s speed and $c$ is the speed of light. This equation demands $v_p$ to be bigger than the speed of light except for light itself or any particle moving with an equivalent speed.
The Klein-Gordon equation $$\frac{1}{c^2}\frac{d^2 \Psi}{dt^2}-\frac{d^2 \Psi}{dx^2}+k_0^2 \Psi=0$$ however, (whose plane waves are also solutions to Dirac’s equation), admits the following plane waves solutions: $$\Psi=e^{i(\sqrt{k^2-k_0^2}x-wt)} \tag{2}$$ Where, $k_0$ is particle's Compton wave-number. The problem here is that this solution is only true if $\frac{w}{k}=c$ as can be easily proven.
So, why is Klein-Gordon’s equation supposed to be valid for a particle having a speed lower than $c$ and simultaneously a phase velocity of $c$?.
PS: I'm aware that particles speed equals the wave's group velocity $v_g$, as can be found from derivating the dispersion relation relatively to $p$. $v_g=\frac{dw}{dk}=\frac{dE}{dp}=\pm c\frac{d}{dp} \sqrt{p^2+(mc)^2}=\frac{c^2p}{E}=v$. So my question is why we have $v_p=c$ if it should be bigger that $c$ as $v=v_g < c$.
