 Solution

Let $P:\mathcal{E}\to \mathcal{D}$ be a discrete fibration and let $m:A\to A^{\prime }$ be monic in $\mathcal{E}$. If $u,v:B\to P(A)$ satisfy $P(m)u=P(m)v$, lift $u,v$ uniquely to $\tilde{u},\tilde{v}$ with codomain $A$. The composites $m\tilde{u},m\tilde{v}$ are lifts of the same arrow with codomain $A^{\prime }$, so uniqueness gives equality. Since $m$ is monic, $\tilde{u}=\tilde{v}$, hence $u=v$. Thus $P(m)$ is monic.

Use the convention that $(F\downarrow B)$ has objects $(A,u:FA\to B)$. Its forgetful functor sends $(A,u)$ to $A$. Given $h:C\to A$, the unique arrow above $h$ with codomain $(A,u)$ has domain $(C,uFh)$. Hence the forgetful functor is a discrete fibration.