= Solution
Consider the commutative square
$$
\begin{array}{ccc}
\mathcal A&\xrightarrow{H}&\mathcal C\\
\downarrow F&&\downarrow G\\
\mathcal B&\xrightarrow{K}&\mathcal D,
\end{array}
$$
where $F$ is a <final functor> and $G$ is a <discrete fibration>. For $B\in\mathcal B$, choose an object $(A,\beta:B\to FA)$ of the nonempty <comma category> $(B\downarrow F)$. Commutativity gives an arrow
$$
K\beta:KB\longrightarrow KFA=GHA.
$$
Lift it uniquely through $G$ with codomain $HA$, and define $LB$ to be the domain of this lift.
This definition does not depend on the choice of $(A,\beta)$. A morphism $u:(A,\beta)\to(A',\beta')$ in the comma category satisfies $Fu\,\beta=\beta'$. The composite of the lift of $K\beta$ with $Hu$ is then a lift of $K\beta'$ with codomain $HA'$, so uniqueness of discrete-fibration lifts says that it is the chosen lift and has the same domain. Since $(B\downarrow F)$ is connected, all choices give the same object $LB$.
For $v:B\to B'$, define $Lv$ as the unique lift through $G$ of $Kv:KB\to KB'=G(LB')$ with codomain $LB'$. Its domain is $LB$: choose $\beta':B'\to FA$, and observe that composing this lift with the lift of $K\beta'$ yields the lift associated with $\beta'v:B\to FA$. Uniqueness of lifting also proves preservation of identities and composition, so $L$ is a functor and $GL=K$.
For $B=FA$, choose $(A,1_{FA})$ in $(FA\downarrow F)$. The lift of $1_{GHA}$ is $1_{HA}$, hence $LFA=HA$; the same lifting argument on arrows gives $LF=H$. Finally, if $L'$ is another filler, then for every $\beta:B\to FA$, the arrow $L'\beta:L'B\to HA$ is a lift of $K\beta$. Unique lifting forces $L'B=LB$ and then forces equality on arrows. Thus $L$ is unique, proving <orthogonality of final functors and discrete fibrations>.
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