$\varphi_\omega$ is a well-defined contraction.

Let $\ell(x)(\omega)\geq0$. From the equality $|\ell(x)|=\ell(|x|)$ we get $\ell(x)(\omega)=\ell(|x|)(\omega)$, which yields $$\varphi_\omega(\ell(x)(\omega))=\varphi_\omega(\ell(|x|)(\omega))=\ell(\mathcal{E}(|x|))(\omega)\geq0,$$ i.e. $\varphi_\omega$ is positive.

The density of $\{\ell(x)(\omega): x\in M\}$ in $L_p(M(\omega),\tau_\omega)$ implies that we can continue $\varphi_\omega$ until a linear positive contraction $T_\omega: L_p(M(\omega),\tau_\omega)\rightarrow
  L_p(N(\omega),\tau_\omega)$ by $$T_\omega(x(\omega))=\lim\limits_{n\rightarrow\infty}\varphi_\omega(\ell(x_n)(\omega))=\lim\limits_{n\rightarrow\infty}\ell(\mathcal{E}(x_n))(\omega).$$

Let $(X,L)$ be a measurable bundle of the noncommutative $L_p(M(\omega),\tau_\omega)$-spaces and $(Y,L)$ be a measurable bundle of the noncommutative $L_p(N(\omega),\tau_\omega)$-spaces. Since $\varphi_\omega(\ell(x)(\omega))\in
\mathcal{L}^\infty(\Omega,Y)$, for any $x\in M$, we obtain that $T_\omega(x(\omega))\in M(\Omega, Y)$ for any $x\in M(\Omega, X)$. Therefore, $\{T_\omega\}$ is a measurable bundle of positive contractions.

It is clear that $\|T_\omega\|_{L_p(M(\omega),\tau_\omega)\rightarrow
L_p(N(\omega),\tau_\omega)}=\|\mathcal{E}\|(\omega)$. From $\|\mathcal{E}\|=\mathbf{1}$, one has $\|T_\omega\|_{L_p(M(\omega),\tau_\omega)\rightarrow
L_p(N(\omega),\tau_\omega)}=1$.

Let us show that $T_\omega$ is a conditional expectation from $M(\omega)$ onto $N(\omega)$.

\(i) Let $x(\omega)\in M(\omega)_{sa}(=\{x(\omega)\in M(\omega):
x(\omega)^*=x(\omega)\}).$ Then $|x(\omega)|\leq
\|x(\omega)\|_{M(\omega)}\id(\omega)$ and $$\begin{aligned}
|T_\omega(x(\omega))|&\leq&
T_\omega(|x(\omega)|)\leq\|x(\omega)\|_{M(\omega)}T_\omega(\id(\omega))\\[2mm]
&=&\|x(\omega)\|_{M(\omega)}\id(\omega),\end{aligned}$$ which means $T_\omega(M(\omega))\subset N(\omega)$ and $T_\omega$ is contraction from $M(\omega)$ to $N(\omega)$, here $\id(\omega)=\ell(\id)(\omega)$.

Let $y(\omega)\in N(\omega).$ Then there exists $y_n\in N$ such that $y(\omega)=\lim\limits_{n\rightarrow\infty}\ell(y_n)(\omega).$ Due to $\mathcal{E}(y_n)=y_n$ we get $$\begin{aligned}
T_\omega(y(\omega))&=&\lim\limits_{n\rightarrow\infty}T_\omega(\ell(y_n)(\omega))=
\lim\limits_{n\rightarrow\infty}\ell(\mathcal{E}(y_n))(\omega)\\[2mm]
&=&\lim\limits_{n\rightarrow\infty}\ell(y_n)(\omega))=y(\omega).\end{aligned}$$ This means $T_\omega(M(\omega))=N(\omega)$, i.e. $T_\omega$ maps $M(\omega)$ onto $N(\omega).$

\(ii) Let $a(\omega), b(\omega)\in N(\omega)$ and $x(\omega)\in
M(\omega)$. Then $$T_\omega
(a(\omega)x(\omega)b(\omega))=\lim\limits_{n\rightarrow\infty}\ell(\mathcal{E}(a_nx_nb_n))(\omega)$$ for some $a_n, b_n\in N$ and $x_n\in M$. Due to $\mathcal{E}(a_nx_nb_n))=a_n\mathcal{E}(x_n)b_n$ we find $$\begin{aligned}
T_\omega
