Kleisli adjunction in a (weak) 2-category In the 2-category of 1-categories, each monad $T$ on a category $\mathcal C$ determines a Kleisli category $\mathcal{C}_T$ and the so-called Kleisli adjunction between categories $\mathcal C$ and $\mathcal{C}_T$.
Let $\mathcal K$ be a (weak) 2-category, $a$ be a $0$-cell in $\mathcal K$, and $t$ be a monad on $a$. Assuming there exists a Kleisli object $a_t$, is there something like a Kleisli adjunction between $a$ and $a_t$? If yes, please describe it.
My attempt following the hint by @KevinCarlson: Assume that there is a Kleisli object $(f_t, \lambda)$. By applying $f_t$ to the identity map of the right $t$-module (that is $(a,t)$), you get a 1-cell from $a_t$ to $a$ that should be the right adjoint. Now, the unit should be $\lambda^{-1} \circ \eta$. But what is the counit? In Cat, the counit $\varepsilon_b$ would simply be $id_{t(b)}$ but I do not see how to generalize that to any 2-category $K$.
 A: Hopefully I can expand on Kevin Arlin's comments in a helpful way.
The answer is yes, there is a Kleisli adjunction.
Preamble
I'll copy the nLab definitions to stay self-contained.
Let $K$ be a 2-category, $t:a\to a$ a monad, $(a_t,f_t,\lambda)$ a Kleisli object for $t$, meaning a representing object for the functor $K\to \newcommand\Cat{\mathbf{Cat}}\Cat\newcommand\oppd{\operatorname{.}}$ that sends an object $x$ to the right $t$-modules on $x$, $\newcommand\RMod{\operatorname{RMod}}\RMod(x,t)$.
So $a_t$ is a 0-cell, $f_t:a\to a_t$ a 1-cell, and $\lambda:f_tt\to f_t$ a 2-cell, such that 
for any right module $(r,\alpha)$, with $r:a\to x$, $\alpha : rt\to r$, there is a unique morphism $a_t\to x$ whose composite with $f_t$ (resp. $\lambda)$ is $r$ (resp. $\alpha$).
Edit: An elementary reformulation of the definition of a Kleisli object:
A Kleisli object for a monad $(a,t:a\to a,\mu:t^2\to t,\eta : 1_a\to t)$ consists of the data of a 0-cell $a_t$, and a right $t$-module $(f_t : a\to a_t,\lambda : f_tt\to f_t)$ on $a_t$ such that the following universality conditions are satisfied.
Object condition: For any right $t$-module on $x$, $(r:a\to x, \alpha : rt\to r)$, there is a unique morphism $g : a_t\to x$ such that 
$(r,\alpha) = (gf_t, g\oppd \lambda)$.
Morphism condition: For two right $t$-modules on $x$, which we know are of the form $(gf_t,g\oppd\lambda)$ and $(hf_t,h\oppd\lambda)$ by the object condition, for $g,h:a_t\to x$ and for every morphism of right $t$-modules $\beta: gf_t\to hf_t$, there is a unique 2-cell $\gamma : g\to h$ such that $\beta = \gamma\oppd f_t$.
The adjoint 1-cells:
We already have $f_t:a\to a_t$, so we need $g_t:a_t\to a$, which should correspond to a right $t$-module structure on $a$. Luckily, we already have a canonical one, $(t,\mu)$, where $\mu:t^2\to t$ is the multiplication of the monad. Thus we get a map $g_t$ from 
the universal property, such that 
$g_tf_t=t$ and $g_t\oppd\lambda = \mu$.
The unit:
Then the unit of the monad, $\eta:1_a\to t=g_tf_t$ is the unit of the adjunction.
Constructing the counit:
To construct the counit, $\epsilon : f_tg_t\to 1_{a_t}$, we need to understand 
$f_tg_t : a_t\to a_t$. However, since $a_t$ represents right modules, this morphism classifies the right module on $a_t$, $(f_tg_tf_t,f_tg_t\oppd\lambda)$, but by definition of 
$g_t$, this is equal to $(f_tt,f_t\oppd\mu)$.
Similarly, $1_{a_t}$ corresponds to the module $(f_t,\lambda)$.
Now you can check that $\lambda: f_tt\to f_t$ is a morphism of right $t$-modules between these two,
since 
$$
\require{AMScd}
\begin{CD}
f_ttt @>f_t\oppd\mu>> f_tt \\
@V\lambda\oppd t VV @VV\lambda V\\
f_tt @>\lambda>> f_t \\ 
\end{CD}
$$
commutes, because this diagram is one of the diagrams that are required for $\lambda$ to be a multiplication making $f_t$ a $t$-module in the first place.
Thus $\lambda$ induces a morphism $\epsilon : f_tg_t\to 1_{a_t}$ satisfying $\epsilon\oppd f_t = \lambda$.
The triangle identities:
For the triangle identities, we now have 
$$(\epsilon\oppd f_t)(f_t\oppd \eta) = \lambda(f_t.\eta)=1_{f_t}$$ by the unit axiom of $\lambda$. For the other, we can understand 
$$(g_t\oppd \epsilon)(\eta \oppd g_t) : g_t\to g_t $$
by composing with $f_t$ to get the corresponding endomorphism of the right $t$-module $(t,\mu)$.
$$((g_t\oppd \epsilon)(\eta\oppd g_t))\oppd f_t
= (g_t\oppd \epsilon \oppd f_t)(\eta\oppd g_t\oppd f_t)
= (g_t\oppd \lambda)(\eta\oppd t)
= \mu(\eta\oppd t)
= 1_t,
$$
by the unit axiom of $\mu$. 
Since $1_t = 1_{g_t}\oppd f_t$, we conclude
$$(g_t\oppd \epsilon)(\eta\oppd g_t) = 1_{g_t},$$
as desired.
