Let $S'$ be the midpoint of $\operatorname{arc} BC$ of $\Omega$, diametrically opposite to $S$ so $SS'$ is a diameter in $\Omega$ and $AS'$ is the angle bisector of $\angle BAC$. Let the tangent of $\omega$ at $P$ meet $\Omega$ again at $Q \neq P$, then we have $\angle SQS' = 90^\circ$.
We will show that triangles $APD$ and $S'QS$ are similar and their corresponding sides are parallel. Then it will follow that the lines connecting the corresponding vertices, namely line $AS'$, that is the angle bisector of $\angle BAC$, line $PQ$, that is the tangent to $\omega$ at $P$, and $DS$ are concurrent. Note that the sides $AD$ and $S'S$ have opposite directions, so the three lines cannot be parallel.

First we show that $AP \perp DP$. Indeed, from cyclic quadrilaterals $APBE$ and $DPLB$ we can see that
$$\angle PAD = \angle PAE = 180^\circ - \angle EBP = \angle PBL = \angle PDL = 90^\circ - \angle ADP.$$
Then, in triangle $APD$ we have $\angle DPA = 180^\circ - \angle PAD - \angle ADP = 90^\circ$.

Now we can see that:
- Both lines $ADE$ and $SS'$ are perpendicular to $BC$, so $AD \parallel S'S$.
- Line $PQ$ is tangent to circle $\omega$ at $P$ so $\angle DPQ = \angle DBP = \angle SBP = \angle SQP$; it follows that $PD \parallel QS$.
- Finally, since $AP \perp PD \parallel QS \perp S'Q$, we have $AP \parallel S'Q$ as well.

Hence the corresponding sides of triangles $APD$ and $S'QS$ are parallel completing the solution.

Again, let $S'$ be the midpoint of arc $BC$, diametrically opposite to $S$, so $AES'S$ is an isosceles trapezoid, and $\angle S'BS = \angle S'PS = 90^\circ$. Let lines $AE$ and $PS'$ meet at $T$ and let $AP$ and $S'B$ meet at point $M$.
We will need that points $L, P, S$ are collinear, and points $T$ and $M$ lie on circle $\omega$.
- From $\angle LPB = \angle LDB = 90^\circ - \angle BDE = 90^\circ - \angle BSS' = \angle SS'B = 180^\circ - \angle BPS$ we get $\angle LPB + \angle BPS = 180^\circ$, so $L, P$ and $S$ are indeed collinear.
- Since $SS'$ is a diameter in $\Omega$, lines $LPS$ and $PTS'$ are perpendicular. We also have $LD \parallel BC \perp AE$ hence $\angle LDT = \angle LPT = 90^\circ$ and therefore $T \in \omega$.
- By $\angle LPM = \angle SPA = \angle SEA = \angle EAS' = \angle EBS' = \angle LBM$, point $M$ is concyclic with $B, P, L$ so $M \in \omega$.

Now let $X$ be the intersection of line $BDS$ with the tangent of $\omega$ at $P$ and apply Pascal's theorem to the degenerate cyclic hexagon $PPMBDT$. This gives points $PP \cap BD = X$, $PM \cap DT = A$ and $MB \cap TP = S'$ are collinear so $X$ lies on line $AS'$, that is the bisector of $\angle BAC$.

Let $A'$ and $S'$ be the points of $\Omega$ diametrically opposite to $A$ and $S$ respectively. It is well-known that $E$ and $A'$ are reflections with respect to $SS'$ so $AS'$ is the angle bisector of $\angle EAA'$. Define point $T$ to be the intersection of $AE$ and $PS'$. As in the previous two solutions, we have: $\angle DPA = 90^\circ$ so $PD$ passes through $A'$; points $L, P, S$ are collinear; and $T \in \omega$.
Let lines $AS'$ and $PDA'$ meet at $R$. From the angles of triangles $PRS'$ and $PTE$ we get
$$\angle ARP = \angle AS'P + \angle S'PA' = \angle AEP + \angle EPS' = \angle ATP$$
so points $A, P, T, R$ are concyclic. Denote their circle by $\gamma$. Due to $\angle RPA = \angle DPA = 90^\circ$, segment $AR$ is a diameter in $\gamma$.

We claim that circles $\omega$ and $\gamma$ are perpendicular. Let line $LPS$ meet $\gamma$ again at $U \neq P$, and consider triangles $PLT$ and $PTU$. By $\angle LPT = \angle TPU = 90^\circ$ and
$$\angle PTL = \angle PBL = 180^\circ - \angle EBP = \angle PAE = \angle PAT = \angle PUT,$$
triangles $PLT$ and $PTU$ are similar. It follows that the spiral similarity that takes $PLT$ to $PTU$, maps $\omega$ to $\gamma$ and the angle of this similarity is $90^\circ$, so circles $\omega$ and $\gamma$ are indeed perpendicular.

Finally, let lines $BDS$ and $ARS'$ meet at $X$. We claim that $X$ bisects $AR$, so point $X$ is the centre of $\gamma$ and, as $\omega$ and $\gamma$ are perpendicular, $PX$ is tangent to $\omega$.
Let $t$ be the tangent of $\omega$ at $D$. From $\angle(DT, t) = \angle TPD = \angle S'PA' = \angle EAS'$ it can be seen that $t \parallel AS'$. Let $I$ be the common point at infinity of $t$ and $AS'$. Moreover, let lines $LPS$ and $ADTE$ meet at $V$. By projecting line $AS'$ to circle $\omega$ through $D$, then projecting $\omega$ to line $AE$ through $L$, finally projecting $AE$ to $\Omega$ through $P$, we find
$$\frac{AX}{RX} = (A, R; X, I) \stackrel{D}{=} (T, P; B, D) \stackrel{L}{=} (T, V; E, D) \stackrel{P}{=} (S', S; E, A') = -1,$$
so $X$ is the midpoint of $AR$.