For notational simplicity let's define

$$ \begin{equation}f_l(\mathbf{q},\sigma):=(\Psi_0,O_l(0)\Psi_{\mathbf{q},\sigma})\end{equation}$$

Now the task is to show $f_l(\mathbf{q},\sigma)$ transforms in the same way as $u_l(\mathbf{q},\sigma)$ under homogeneous Lorentz transformation $\Lambda$, i.e. transforms as given in equation (5.1.19), which, together with irreducibility, defines $u_l(\mathbf{q},\sigma)$ uniquely up to a constant multiplication.

Due to the Lorentz invariance of vacuum and the fact $\Lambda 0=0$, we have(adopting the repeated index summation convention)

$$\begin{equation}f_m(\mathbf{q},\sigma)=(\Psi_0,U(\Lambda)O_m(0)U^{-1}(\Lambda)U(\Lambda)\Psi_{\mathbf{q},\sigma})\\=D_{ml}^{-1}(\Lambda)(\Psi_0,O_{l}(0)U(\Lambda)\Psi_{\mathbf{q},\sigma}),\end{equation}$$

which is equivalent to

$$\begin{equation}D_{lm}(\Lambda)f_m(\mathbf{q},\sigma)=(\Psi_0,O_{l}(0)U(\Lambda)\Psi_{\mathbf{q},\sigma}).\end{equation}$$

Now substitute equation (2.5.11) to the RHS we get

$$\begin{equation} D_{lm}(\Lambda)f_m(\mathbf{q},\sigma)=\sqrt{\frac{(\Lambda p)^0}{p^0}}D^{(j_n)}_{\sigma'\sigma}(W)(\Psi_0,O_{l}(0)\Psi_{\mathbf{q}_\Lambda,\sigma'})\\=\sqrt{\frac{(\Lambda p)^0}{p^0}}D^{(j_n)}_{\sigma'\sigma}(W)f_l(\mathbf{q}_\Lambda, \sigma')\end{equation}$$

This is exactly (5.1.19).

Note that $D_{lm}(\Lambda)$ is assumed to be irreducible in the book,

...where $O_l(x)$ is a Heisenberg-picture operator, with the Lorentz transformation properties of some sort of free field $\psi_l$ belonging to an irreducible representation of the homogeneous Lorentz group... (page 437)

hence by the discussion in chapter 5(e.g. content at the bottom of page 196), our function can differ $u_l(\mathbf{q},\sigma)$ only by a constant multiplication.