The sets $A$ and $B$ that I constructed in that earlier answer don’t solve the original problem: that question asked how to show Ken’s hint. The hint is this:
$\mathcal{A}=\{a_\alpha:\alpha<\omega_1\}$, and $\mathcal{B}\setminus\mathcal{A}=\{b_\alpha:\alpha<\omega_1\}$. Construct $a_\alpha,b_\alpha$ inductively so that $a_\alpha\cap b_\alpha=\varnothing$ but $\alpha\ne\beta\to a_\alpha\cap b_\beta\ne\varnothing$.
The construction of $A$ and $B$ in that earlier answer is how you construct $a_\alpha$ and $b_\beta$ when you’ve already constructed $a_\xi$ and $b_\xi$ for $\xi<\alpha$. The $\mathscr{A}$ in that answer is $\{a_\xi:\xi<\alpha\}$, temporarily re-indexed by $\omega$, and the $\mathscr{B}$ is $\{b_\xi:\xi<\alpha\}$. To finish the proof once you’ve constructed the subsets $a_\alpha$ and $b_\alpha$ of $\omega$, let $$\mathcal{A}=\{a_\alpha:\alpha<\omega_1\}$$ and $$\mathcal{B}=\mathcal{A}\cup\{b_\alpha:\alpha<\omega_1\}\;.$$ By construction these are almost disjoint families, and $|\mathcal{A}|=|\mathcal{B}\setminus\mathcal{A}|=\omega_1$, but we still need to show that they satisfy the requirements of the original problem.
Unfortunately, that problem seems to be misstated. In my copy of Ken’s book it reads as follows:
Show (in $\mathbf{ZFC}$) that the result of Exercise $9$ can be false if $|\mathscr{A}|=|\mathscr{B}\setminus\mathscr{A}|=\omega_1$.
(This is then followed by the hint.) The result from Exercise $9$ is that if $\mathscr{B}\subseteq\wp(\omega)$ is an almost disjoint family of size $\omega$, and $\mathscr{A}\subseteq\mathscr{B}$ is countable, then there is a $d\subseteq\omega$ such that $\forall x\in\mathscr{A}(|d\cap x|<\omega)$ and $\forall x\in\mathscr{B}\setminus\mathscr{A}(|x\setminus d|\color{red}{<}\omega)$. Thus, we need to show that there is no $d\subseteq\omega$ such that $|d\cap a_\alpha|<\omega$ and $|b_\alpha\setminus d|<\omega$ for each $\alpha<\omega_1$. (That is, no subset of $\omega$ is almost disjoint from each $a_\alpha$ and almost contains each $b_\alpha$.)
Suppose, on the contrary, that $d$ is such a set. Now $\omega$ has only $\omega$ finite subsets, so there are finite $r,s\subseteq\omega$ and an uncountable $A\subseteq\omega_1$ such that $d\cap a_\alpha=r$ and $b_\alpha\setminus d=s$ for each $\alpha\in A$. Recall that $a_\alpha\cap b_\alpha=\varnothing$, so $r\cap s=\varnothing$. Suppose that $\alpha,\beta\in A$, and $\alpha\ne\beta$; then $a_\alpha\cap b_\beta\ne\varnothing$, so let $n\in a_\alpha\cap b_\beta$.
If $n\in d$, then $n\in a_\alpha\cap d=r=a_\beta\cap d$; but then $n\in a_\beta\cap b_\beta=\varnothing$, which is absurd, so $n\notin d$, and therefore $n\in b_\beta\setminus d=s$. And that is also impossible: then $n\in s=b_\alpha\setminus d$ and hence $n\in a_\alpha\cap b_\alpha=\varnothing$. Thus, there is indeed no such $d\subseteq\omega$.
