This is basically a long comment to your answer saying a bit more about the structure of $\mathbb A(\kappa)$. The forcing $\mathbb A(\kappa)$ is equivalent to $\mathrm{Add}(\kappa, 1)\ast\dot{\mathbb P}$ where $\dot{\mathbb P}$ is forced to have the following properties:

 - It is $\kappa$-cc of size $\kappa$,
 - collapses all cardinals below $\kappa$ and
 - is not isomorphic to any forcing in $V$, in particular it is not the Levy collapse $\mathrm{Col}(\omega,{<}\kappa)$. 

Let me sketch how to see this. Let $\mathbb Q$ be the forcing consisting of only the first components of $\mathbb{A}(\kappa)$ with the order inherited in the obvious way. $\mathbb Q$ adds a directed system of forcings so that the forcings appearing earlier in the system are regular subforcings of the later ones. We have that forcing with $\mathbb A(\kappa)$ is equivalent to $\mathbb Q\ast\dot{\mathbb P}$ where $\dot{\mathbb{P}}$ is a name for the direct limit of the system added by $\mathbb Q$. Further, $\mathbb Q$ is a nonatomic ${<}\kappa$-closed forcing of size $\kappa$ and hence equivalent to $\mathrm{Add}(\kappa, 1)$.

Now let $G$ be $\mathbb{Q}$-generic and  $\mathbb P=\dot{\mathbb{P}}^G$. Clearly $\mathbb P$ is of size $\kappa$ and the argument in your answer shows that $\mathbb P$ is $\kappa$-cc: If $\dot A$ is a $\mathbb Q$-name for a maximal antichain in $\dot{\mathbb P}$ then there is a forcing $\mathbb R\in G$ so that 
$$A_0:=\{r\in\mathbb R\mid \mathbb R\Vdash_{\mathbb Q}\check r\in\dot A\}$$
is a maximal antichain in $\mathbb R$. Now $\dot A$ cannot grow any larger later, so $\dot A^G=A_0$ is small. 

As you note, $\mathbb A(\kappa)$ collapses all cardinals below $\kappa$, the same must be true for $\mathbb P$ (which can also be seen in the same way directly).

Finally, suppose toward a contradiction that $\mathbb P$ is isomorphic to some forcing $\mathbb P'$ in $V$. By nature of how $\mathbb P$ arises, we can find a sequence $\vec {\mathbb P}:=\langle \mathbb P_\alpha\mid\alpha<\kappa\rangle$ which satisfies

 - $\mathbb P=\bigcup_{\alpha<\kappa}\mathbb P_\alpha$,
 - all $\mathbb P_\alpha$ are of size ${<}\kappa$ and
 - $\mathbb P_\alpha\lessdot\mathbb P_\beta\lessdot \mathbb P$ whenever $\alpha<\beta<\kappa$ ($\lessdot$ denotes regular subforcing).

We can find (in $V$!) a sequence $\vec{\mathbb P}'=\langle\mathbb P_\alpha'\mid\alpha<\kappa\rangle$ with analogous properties relative to $\mathbb P'$. 
Now consider 
$$\Delta\left(\vec{\mathbb P}\right)=\left\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb P_{\beta}\lessdot\mathbb P\right\}.$$

This set modulo $\mathrm{NS}_\kappa$ does not depend on the particular choice of $\vec{\mathbb P}$. It thus suffices to show that $\Delta\left(\vec{\mathbb P}\right)\neq\Delta\left(\vec{\mathbb P}'\right)\mod\mathrm{NS}_\kappa$ and for this it suffices to show that $\Delta\left(\vec{\mathbb P}\right)$ splits every stationary subset of $\kappa$ in $V$ into two stationary sets. The main idea here is that whenever $(\mathbb R_\alpha)_{\alpha<\gamma}$ is a strictly decreasing sequence in $\mathbb Q$ of length $\gamma<\kappa$ then both the direct limit $\mathbb R_{\mathrm{dir}}$ and the inverse limit $\mathbb R_{\mathrm{inv}}$ along this sequence produce lower bounds in $\mathbb Q$. However, we have that $\bigcup_{\alpha<\gamma}\mathbb R_\alpha$ is not a regular subforcing of $\mathbb R_{\mathrm{inv}}$ but it is of $\mathbb R_{\mathrm{dir}}$ (in fact this is $\mathbb R_{\mathrm{dir}}$). This in turn decides whether or not $\bigcup_{\alpha<\gamma}\mathbb R_\alpha\lessdot\mathbb P$.