 Solution

The boundary arcs meet at $0$ and $1$. The Möbius transformationsends these points to $0$ and infinity, sends the real boundary segment to the positive real axis, and sends the circular boundary arc to the ray of argument $-\pi /6$. Hence it maps $D\cap L$ onto the sectorThe power $\zeta =w^6$ maps this sector conformally onto the lower half-plane. Finally, the Cayley map $\zeta \mapsto (\zeta +i)/(\zeta -i)$ maps the lower half-plane onto the unit disc. Thus a requested conformal equivalence is