= Solution
{c}
For $L\dashv R$ inducing $T$, the <Kleisli comparison functor> is always <full and faithful>, since the adjunction gives
$$
\mathcal D(LA,LB)\cong\mathcal C(A,RLB)=\mathcal C_T(A,B).
$$
Therefore it is part of an <equivalence of categories> if and only if every object of $\mathcal D$ is isomorphic to $LA$ for some $A$.
For an <idempotent monad>, multiplication is invertible. The unit laws imply $T\eta=\eta T=\mu^{-1}$. If $(A,a)$ is an <algebra for a monad>, then $a\eta_A=1_A$, while naturality of $\eta$ gives
$$
\eta_Aa=T(a)\eta_{TA}=T(a)T(\eta_A)=T(a\eta_A)=1_{TA}.
$$
Thus $a:TA\to A$ is an isomorphism with inverse $\eta_A$. The algebra law $a\mu_A=aT(a)$ says that $a$ is a <morphism of algebras for a monad> from the free algebra $(TA,\mu_A)$ to $(A,a)$, so every object of the <Eilenberg-Moore category> is isomorphic to a free algebra. The criterion above yields
$$
\boxed{\mu\text{ invertible}\ \Longrightarrow\ \mathcal C_T\simeq\mathcal C^T.}
$$
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