To prove $\measuredangle AKL = \dfrac{\measuredangle ABC}{2}$ (see in $\mathrm{Fig.\space 2}$), we need the following simple lemma.
$\bf{Lemma}:$
As shown in $\mathrm{Fig.\space 1}$, Points $E$ and $F$ are taken on the extended sides $AD$ and $BC$ respectively of the cyclic quadrilateral $ABCD$, such that its side $CD$ is parallel to the segment $EF$. Show that the quadrilateral $ABEF$ is cyclic.
Before reading our proof hidden in the spoiler box below, we would like to see OP trying to prove it him/her-self.
$\bf{Proof\enspace of\enspace Lemma}:$
Let $\measuredangle ABF = \psi$. Since the external angle at a vertex of a cyclic quadrilateral is equal to the internal angle at its opposite vertex, we have,
$$\measuredangle EDC = \measuredangle ABF = \psi. \tag{1}$$
Furthermore, Since $CD$ is parallel to $EF$, we shall write,
$$\measuredangle AEF = \measuredangle EDC = \psi. \tag{2}$$
It is evident from (1) and (2) that $\measuredangle AEF= \measuredangle ABF$. This means that the straight line $AF$ subtends equal angle at $B$ and $E$. According to the converse of the theorem Euclid-III-21, $ABEF$ is a cyclic quadrilateral.
$\mathbf{Proof\enspace of\enspace Parallelity \enspace of} \enspace \pmb{AC}\enspace \mathbf{and}\enspace \pmb{PQ} :$
As shown in $\mathrm{Fig.\space 2}$, segments $AK$ and $BL$ intersect with each other at $P$, while segments $B_1K$ and $BC$ at $Q$. We add the line segments $BK$, $MA$, and $PQ$ to the original configuration described by OP in the problem statement. Furthermore, for brevity, let $\measuredangle ABC = 2\phi$, $\measuredangle BCA = \alpha$, and $\measuredangle LBC= \beta$.
Since $\measuredangle BKA$ and $\measuredangle BCA$ are in the same segment of the circle $\omega$, we have,
$$\measuredangle BKA = \measuredangle BCA = \alpha. \tag{3}$$
Since $\measuredangle KPB =\measuredangle KQB = 90^o$, according to the converse of the theorem Euclid-III-21, the quadrilateral $BKQP$ is cyclic. This means that $\measuredangle BKP$ and $\measuredangle BQP$ are in the same segment of the circle $BKQP$. Therefore, using (3), we shall state,
$$ \measuredangle BQP = \measuredangle BKA =\alpha.$$
This means that $\measuredangle BQP = \measuredangle BCA$, and, hence, $PQ$ is parallel to $CA$.
$\mathbf{Proof\enspace of}\enspace \pmb{\measuredangle AKL = \dfrac{\measuredangle ABC}{2}} :$
The points $B_1$ and $L$ lies on the extended sides $KQ$ and $BP$ respectively of the cyclic quadrilateral $BKQP$. According to the lemma proven above, $BKLB_1$ is also a cyclic quadrilateral.
Since $BB_1$ is the angle bisector of $\measuredangle ABC$, we have $\measuredangle B_1BC = \phi$. With this we can determine $\measuredangle B_1BL$ as shown below.
$$\measuredangle B_1BL = \measuredangle B_1BC - \measuredangle LBC = \phi - \beta$$
Since $BKLB_1$ is a cyclic quadrilateral, according to the theorem Euclid-III-21,
$$\measuredangle B_1KL = \measuredangle B_1BL = \phi - \beta. \tag{4}$$
In a similar vein, since $BKQP$ is a cyclic quadrilateral, according to the theorem Euclid-III-21,
$$\measuredangle PKQ= \measuredangle PBQ = \beta. \tag{5}$$
Using (4) and (5), we can express the magnitude of the $\measuredangle AKL$ as
$$\measuredangle AKL = \measuredangle PKQ + \measuredangle B_1KL = \phi. \tag{6}$$
We describe below how to prove that $K$, $L$, and $M$ are collinear. Here too, we encourage OP to work out own proof before reading ours.
$\mathbf{Proof\enspace of\enspace collinearity\enspace of\enspace} \pmb{K},\enspace \pmb{L},\enspace \mathbf{ and }\enspace \pmb{M} :$
Since $\measuredangle AKM$ and $\measuredangle ABM$ are in the same segment of the circle $\omega$, we have,
$$ \measuredangle AKM = \measuredangle ABM = \phi. \tag{7}$$
It follows from (6) and (7) that
$$\measuredangle AKM = \measuredangle AKL.$$
This can be possible if and only if $K$, $L$, and $M$ are collinear.
Best Answer
Proving that $PQ$ is tangent to $(O)$ is enough.
Here, I shall solve a subproblem and then use that to solve the original.
Subproblem :
Let $AB$ be a chord of $(O)$. $A'$ is $A$ reflected about $O$. $D$ is a point on $AB$. The perpendicular to $AB$ at $D$ meets the tangent to $(O)$ at $A$, at point $E$. $OD$, when extended, meets the tangent to $(O)$ at $B$, at point $C$.
Claim: $E$, $C$ and $A'$ are collinear.
Proof:
Let $CA'$ intersect $(O)$ at $F$. $M$ is the feet of the perpendicular from $B$ on $CD$. $O$, $M$, $F$ and $A'$ are concyclic since $CM \cdot CO= CB^2= CF \cdot CA'$. So, $\angle MFB= \angle MFA' + \angle A'FB= \angle AOD + \angle OAD= 180^{\circ}- \angle MDB$ and thus $DMFB$ is cyclic.
$\implies \angle DFB= 90^{\circ}$.
Now, $\angle DFA= \angle A'FB= \angle A'AB= \angle AED$ and hence $DAEF$ is cyclic.
$\implies \angle AFE= 90^{\circ}$ and so, $A'$, $C$ and $E$ are collinear (from collinearity of $A'$, $F$ and $E$).
In the actual Problem:
Let $Q'$ be the point where the tangent to $(O)$ intersects $OD$(extended). From earlier, points $S$, $Q'$ and $K$ are collinear and so, $Q'=Q$. Therefore, $PQ$ is tangent to $(O)$.