= Solution
In the state-picture convention, a <Stinespring representation of a completely positive map> $\mathcal N:\mathcal L(\mathcal H_Q)\to\mathcal L(\mathcal H_{Q'})$ consists of an auxiliary <Hilbert space> $\mathcal H_E$ and a linear map $V:\mathcal H_Q\to\mathcal H_{Q'}\otimes\mathcal H_E$ such that
$$
\boxed{\mathcal N(X)=\operatorname{Tr}_E(VXV^\dagger)\quad\text{for all }X.}
$$
The <partial trace> discards the environment. For example, from a <Kraus representation> $\mathcal N(X)=\sum_jK_jXK_j^\dagger$, take $V=\sum_jK_j\otimes|j\rangle_E$. The adjoint, or observable-picture, form is $\mathcal N^\dagger(Y)=V^\dagger(Y\otimes I_E)V$.
A general <completely positive map> does not require $V$ to be an <linear isometry of Hilbert spaces>. If $\mathcal N$ is trace preserving, then $V^\dagger V=\sum_jK_j^\dagger K_j=I$, so $V$ is an <linear isometry of Hilbert spaces>. This is the <Stinespring dilation> of a <quantum channel>, which will be used in the data-processing proof.
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