The denominator, $|\vec{p}|$, is the magnitude of the three-momentum $\vec{p}$:
$$|\vec{p}|=\sqrt{p_x^2+p_y^2+p_z^2}$$
This is the momentum you know from classical physics, and it is different from the momentum operator
$$\hat{\textbf{p}} = - i \hbar \frac{\partial}{\partial x},$$
or the four-momentum $p_{\mu}=(E,p_x,p_y,p_z)$.
Now, helicity is defined as
$$h= \frac{\vec{S}\cdot\vec{p}}{|\vec{p}|},$$
which is the normalized component of the particle's spin along its direction of flight.
Here is the deal: there are four independent basis states for solutions of the Dirac equation. Two of these correspond to particle solutions and two to antiparticle solutions. The question is, why do we have two of each and what is the difference between them?
The helicity operator, defined as
$$\hat{h} = \frac{{\hat{\textbf{Σ}}} \cdot {\hat{\textbf{p}}}}{|\vec{p}|}= \frac{1}{2|\vec{p}|}\begin{pmatrix}
\textbf{σ} \cdot \hat{\textbf{p}} & 0\\
0 & \textbf{σ} \cdot \hat{\textbf{p}}
\end{pmatrix},$$
measures how the particle’s spin axis is aligned with the particle’s motion: whether it is parallel or antiparallel. More importantly, it commutes with the free particle Dirac Hamiltonian and is therefore a conserved quantity (but not Lorentz invariant!). Since they commute, it is possible to identify spinor states which are simultaneous eigenstates of both operators ($\hat{H}_D$ and $\hat{h}$). For a spin 1/2 particle, the component of spin measured along any axis is quantised to be either $\pm 1/2$. Therefore, acting the helicity operator on the Dirac spinors gives the eigenvalues $\pm 1/2$. The two possible helicity states for a spin 1/2 fermion are termed right-handed and left-handed helicity states. This is why we have four independent basis states, two for particle and two for antiparticle solutions.
P.S. I'm pretty new to this so take what I say with a grain of salt. The figure is taken from Modern Particle Physics by Mark Thomson.
