In Exercise (ECMP) (p.91-92) of the book Toposes, Triples, and Theories, the reader is asked to prove that the following diagram
is a pullback, where $T$ is a monad on $\mathcal{C}$, $\mathcal{K}=\mathcal{C}_T$ is the Kleisli category (regarded as the full subcategory of $\mathcal{C}^T$ consisting of all free $T$-algebras as objects), $U^T$ is the forgetful functor, $F_T:\mathcal{C} \to \mathcal{K}$ is the free $T$-algebra functor, and $H$ is the functor
$$(A,h) \mapsto \mathcal{C}^T(-,(A,h))|_\mathcal{K}: \mathcal{K}^\mathrm{op} \to \mathbf{Set}$$
However, the exercise seems to make no sense, and here is why. Take a $T$-algebra $(A,h)$. Then the lower path yields the representable $\mathcal{C}(-,A)$ and the upper path yields the functor $\mathcal{C}^T(F_T^\mathrm{op}(-),(A,h))$. The diagram commute precisely means these two functors are the same. Let $B \in \mathcal{C}$. Then we need to show that $$\mathcal{C}(B,A) = \mathcal{C}^T(F_T^\mathrm{op}(B),(A,h)) = \mathcal{C}^T((TB,\mu_B),(A,h))$$ But this doesn't make sense, because the LHS consists of maps $B \to A$ while the RHS consists of maps $TB \to A$ (and compatible with the structure maps). My guess is that I (somehow) misunderstand the exercise, but not sure where.
Next, suppose this diagram commute (somehow), and $M:\mathcal{D} \to \mathcal{C}$ and $N:\mathcal{D} \to \mathbf{Func}(\mathcal{K}^\mathrm{op},\mathbf{Set})$ be functors with $$\text{Yoneda} \circ M = \mathbf{Func}(F_T^\mathrm{op},\mathbf{Set}) \circ N.$$ To prove the universal property, we need a unique functor $P: \mathcal{D} \to \mathcal{C}^T$ such that $M=U^T \circ P$ and $N=H \circ P$. The functor $P$ must sends $D \in \mathcal{D}$ to a $T$-algebra $(MD,h)$ for some structure map $h:TMD \to MD$. But I have no clue on what should the map $h$ be. I guess it has something to do with $N$ but unable to figure that out.
Any help is appreciate.