 Solution

The spin operators here are dimensionless: $S^j={\sigma }_j/2$, so $\Omega$ has units of inverse time and $\Phi$ is a dimensionless pulse area. Integrating the Schrödinger evolution across an infinitesimally short pulse gives $U_{\Phi }=e^{-i\Phi S^x}$. The finite free Hamiltonian contributes a vanishing integrated phase within that pulse. Between pulses the free evolution is $U_{\Omega }=e^{-i\Omega T{S}^z}$.

In the $S^z$ eigenbasis, $S^z=\mathrm{diag}(1,-1)/2$ and $({\sigma }_x{)}^2=I$. ConsequentlyStarting just before a pulse, the kick acts first and the free interval second. Operators act on states right to left, henceThere are exactly $N$ kicks, at times $0,T,\dots ,(N-1)T$; the pulse at $NT$ is excluded. Sampling just after kicks would change the displayed product order. This endpoint convention specifies the delta-kicked spin Floquet operator.