The question uses notation that doesn't distinguish between covariant and contravariant functors, which I think is confusing. Here's the result restated with $^\mathrm{op}$'s in the right places: $\mathrm{Lan}_{K^\mathrm{op}} N(-/\mathcal{C}) \cong N(-/K)$ where $N(-/\mathcal{C}) : \mathcal{C}^\mathrm{op} \to \mathbf{Cat}$, $N(-/K) : \mathcal{D}^\mathrm{op} \to \mathbf{Cat}$ and $K^\mathrm{op}$ is the same functor as $K$ but regarded as a functor $\mathcal{C}^\mathrm{op} \to \mathcal{D}^\mathrm{op}$.
All right, let's compute the left Kan extension. Its value at $d^\mathrm{op} \in \mathcal{D}^\mathrm{op}$ is given by $\mathop{\mathrm{colim}}\left( \Phi : K^\mathrm{op}/d^\mathrm{op} \to \mathcal{C}^\mathrm{op} \xrightarrow{N(-/\mathcal{C})} \mathbf{sSet} \right)$. Under the isomorphism $K^\mathrm{op}/d^\mathrm{op} \cong (d/K)^\mathrm{op}$, $\Phi : (d/K)^\mathrm{op} \to \mathbf{sSet}$ is given by $\Phi(\alpha : d \to Kc) = N(c/\mathcal{C})$. (Notice that this really is contravariant.)
Now, the trick is to notice that for $\alpha : d \to Kc$, the category $c/\mathcal{C}$ is isomorphic to $\alpha/(d/K)$, so that the functor $\Phi$ is of the form $N(-/\mathcal{A}) : \mathcal{A}^\mathrm{op} \to \mathbf{sSet}$ for $\mathcal{A} = d/K$ and we must show $\mathop{\mathrm{colim}} N(-/\mathcal{A}) = N(\mathcal{A})$.
Remark 1: We've essentially reduced your question to the case $\mathcal{D} = 1$.
Remark 2: What's left to show is equivalent to the fact that the homotopy colimit of the constant functor with value a point is the nerve of the indexing category, or, more precisely, that the Bousfield-Kan formula (with appropriate choice of op/no-op) for that colimit gives exactly the nerve.
One can prove the remaining claim directly: the colimiting cocone $\pi : N(-/\mathcal{A}) \Rightarrow N(\mathcal{A})$ is given by applying the nerve functor to the canonical projections $a/\mathcal{A} \to \mathcal{A}$; given any simplicial set $B$ and any cocone $\gamma : N(-/\mathcal{A}) \Rightarrow B$, we'll prove it factors through $\pi$ by means of the map $G : N(\mathcal{A}) \to B$ given by $G(\sigma) = \gamma_{a_0}(\bar\sigma)$, where $\sigma$ is a simplex $a_0 \xrightarrow{f_0} \cdots \xrightarrow{f_{n-1}} a_n$ of $N(\mathcal{A})$ and $\bar\sigma$ is the corresponding simplex $1_a \to (a_0 \xrightarrow{f_0} a_1) \to (a_0 \xrightarrow{f_1 f_0} a_2) \to \cdots (a_0 \xrightarrow{f_{n-1} \cdots f_0} a_n)$ of $N(a_0/\mathcal{A})$.
To check $G$ is a map of simplicial sets, we use the naturality of $\gamma$. Let $h : [m] \to [n]$ be a simplicial operator and $\sigma = (a_0 \xrightarrow{f_0} \cdots \xrightarrow{f_{n-1}} a_n)$ an $n$-simplex of $N(\mathcal{A})$. If $h(0) = 0$, then $G(h^\ast \sigma) = h^\ast G(\sigma)$ just follows from $\gamma_{a_0}$ being a map of simplicial sets. If instead $h(0) > 0$, let $f = f_{h(0)-1} \cdots f_0 : a_0 \to a_{h(0)}$. By naturality of $\gamma$, $\gamma_{a_{h(0)}} = \gamma_{a_0} \circ f^\ast$. It's easy to check by hand that $f^\ast (\overline{h^\ast \sigma}) = h^\ast(\bar\sigma)$, so we get $$G(h^\ast \sigma) = \gamma_{a_{h(0)}}(\overline{h^\ast \sigma}) = \gamma_{a_0} (f^\ast (\overline{h^\ast \sigma})) = \gamma_{a_0} (h^\ast(\bar\sigma)) = h^\ast (\gamma_{a_0}(\bar\sigma)) = h^\ast(G(\sigma)).$$
An $n$-simplex $\tau \in N(a/\mathcal{A})$ is determined by the $n$-simplex $\pi_a(\tau)$ in $N(\mathcal{A})$ and the morphism $f : a \to a_0$ where $a_0$ is the first vertex of $\pi_a(\tau)$. With that notation we can easily establish that $G$ is the unique map such that $G \cdot \pi = \gamma$:
Let's check that $G(\pi_a(\tau)) = \gamma_a(\tau)$: we have
$f^\ast(\overline{\pi_a(\tau)}) = \tau$, so $\gamma_a(\tau) =
\gamma_a(f^\ast(\overline{\pi_a(\tau)})) =
\gamma_{a_0}(\overline{\pi_a(\tau)}) = G(\pi_a(\tau))$.
Now to see that $G$ is the unique such map, take any $n$-simplex $\sigma$ in $N(\mathcal{A})$ and let $\tau = \bar\sigma$. Then $\pi_{a_0} \tau = \sigma$ (where $a_0$ is the initial vertex of $\sigma$) and thus $G(\sigma) = G(\pi_{a_0}(\tau)) = \gamma_{a_0}(\bar\sigma)$ is forced.
