= Solution
Choose the doublet <bases> so that the column of barred fields $B=(\bar\psi_L^1,\bar\psi_L^2)^T$ and $R=(\psi_R^1,\psi_R^2)^T$ both transform by the doublet <matrices> in <1/d>. This choice follows the source's assignment of the barred doublet itself. Since $H$ transforms in $\varepsilon$, its <Yukawa coupling> must multiply a fermion bilinear in $\varepsilon$.
There is exactly one such bilinear in $\mathbf2\otimes\mathbf2$: the <Clebsch-Gordan coefficients> in <1/d> give $B_1R_2-B_2R_1$. Equivalently, writing the upper block of the <Yukawa matrix> as $A$, invariance requires
$$
D(r)^TAD(r)=A,\qquad D(s)^TAD(s)=-A.
$$
The first equation sets $A_{11}=A_{22}=0$, because $\alpha^2\ne1$; the second sets $A_{21}=-A_{12}$.
A mixed doublet-singlet bilinear remains a doublet after multiplication by $H$, so all four mixed entries vanish. Finally $\bar\psi_L^3\psi_R^3$ transforms in $\varepsilon\otimes\mathbf1=\varepsilon$, and multiplication by $H$ makes it invariant. Therefore
$$
\boxed{\lambda=\begin{pmatrix}0&a&0\\-a&0&0\\0&0&b\end{pmatrix},\qquad a,b\in\mathbb C.}
$$
The allowed <Yukawa interaction> is $H\{a(\bar\psi_L^1\psi_R^2-\bar\psi_L^2\psi_R^1)+b\bar\psi_L^3\psi_R^3\}$, together with its <Hermitian conjugate> when forming a real <Lagrangian density>.
The displayed texture depends on <basis>. If instead the unbarred left doublet is assigned $D$ in the same ordering as the right doublet, its barred doublet has the <dual representation> $D^*$ and the equivalent upper block is $a\operatorname{diag}(1,-1)$. Exchanging the two barred components converts between these descriptions. In either convention there are exactly two independent complex <Yukawa couplings>.
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