The particle speed $v=\sqrt{\dot y^2+\dot z^2}$.

$\dot y=-\omega a \cos\phi$

$\dot z=+\omega a\sin\phi$

where $\phi=kx-\omega t$.

So although $\dot y, \dot z$ each depend on $x, t$ the combination $v=\sqrt{\dot y^2+\dot z^2}=\omega a=$ constant does not depend on either $x$ or $t$.