Let the perpendicular bisector of the segment $ TX $ intersect $XK$ at $Y$ and $AY \cap XB =R$. Show that $R$ is on $(APQB)$ and the rest is easy. I will add the details, motivation etc. later.

Let $AK\cap (ABPQ)=\{A,S\}$. As $\measuredangle SBK=\measuredangle SAB=\measuredangle BLK=\measuredangle LBK$, we find that $BL$ passes through $S$. We have $\angle XBQ+\angle BXQ=\angle BQP=\angle BAP=\angle BAK+\angle SAP=\angle BXQ+\angle LBP$, yields $\angle XBQ=\angle LBP$. Hence, $P$ and $Q$ are isogonal pairs wrt $\triangle XBL$. We know $A$ is the center of the spiral symmetry that takes $XL$ to $TB$. Yields, $\triangle AXL\sim\triangle ATB\Rightarrow |TB|=\frac{|XL||AB|}{|AL|}=\frac{|XL||BS|}{|SL|}.$ Let $XK\cap BL=Y$. Now, we will prove that $\angle TQB=\angle LQY$. By the law of Sines in the triangle $TQB$, we get $\frac{\sin(BQT)}{\sin(BQT+TBQ)}=\frac{|TB|}{|QB|}$. We have $\angle SQY=\angle SQP=\angle SBP=\angle TBQ$. By comparing the law of Sines in the triangles $LQY$ and $LQS$, we get $\frac{\sin(LQY)}{\sin(LQY+TBQ)}=\frac{\sin(LQY)}{\sin(LQY+YQS)}=\frac{|QS||LY|}{|QY||LS|}$. Hence, in order to prove that $\angle TQB=\angle LQY$, we should show that $\frac{|QS||LY|}{|QY||LS|}\overset{?}{=}\frac{|TB|}{|QB|}=\frac{|XL||BS|}{|SL||QB|}\Leftrightarrow \frac{|QS||QB|}{|QY||BS|}\overset{?}{=}\frac{|XL|}{|LY|}.$ We have $\angle BQS=\angle BAK=\angle BXQ=\angle LXY$. By comparing the law of Sines in the triangles $BQS$ and $BQY$, we get $\frac{|QS||QB|}{|QY||BS|}=\frac{\sin(BQS)}{\sin(BYQ)}=\frac{\sin(LXY)}{\sin(XYL)}$. Hence, it suffices to prove that $\frac{|XL|}{|LY|}\overset{?}{=}\frac{\sin(LXY)}{\sin(XYL)}$, which is evident by the law of Sines in the triangle $XYL$. Thus, we showed that $\angle TQB=\angle LQY$. Similarly, one can show that $\angle TPB=\angle LPY$. From here, we can get the desired result by simple angle-chasing.

Let $BX$ meet $AQP$ at $R$. Let $AR$ meet $XK$ at $S$. Claim $: ATSX$ is cyclic. Proof $:$ $\angle TAB = \angle TBL = \angle XAL \implies \angle XAT = \angle LAB = 2\angle TXS$ and also $\angle ARB = \angle 180 - \angle ABK = \angle AXK \implies \angle RAX = \angle RXS$ so now we have $\angle TAS = \angle TAX - \angle RAX = 2\angle TXS - \angle TXS = \angle TXS$ which implies $ATSX$ is cyclic. Now we'll prove $\angle PTX = \angle QTB$ which will end the proof. Note that $\angle PTX = \angle 180 - \angle TPQ - \angle TXS$ and $\angle QTB = \angle 180 - \angle XBP - \angle TQB$ so we need to prove $\angle TPQ + \angle TXS = \angle XBP + \angle TQB$. we add $\angle TXS$ to both sides and now we need to prove $\angle TPQ + \angle BXL = \angle BQP + \angle TQB = \angle TQP$. Note that $\angle TQP = \angle TSQ + \angle STQ = \angle TAX + \angle STQ = \angle BAL + \angle STQ = \angle BXL + \angle STQ$ so now we need to prove $\angle TPQ = \angle STQ$ which is equivalent to $ST^2 = SQ.SP$. Note that $SQ.SP = SR.SA$ and $\angle STR = \angle SAX = \angle SXR = \angle SXT = \angle SAT$ so $ST^2 = SR.SA$ as wanted. we're Done.

This problem is very beautiful, would make a very good IMO 2/5. Initially, we notice that $\triangle ATX \sim \triangle ABL$ which is not very useful on its own. However, the fact that $\angle ATX = \frac{\angle BXK}{2}$ is quite interesting because it means that if we introduce $K'$ to be the midpoint of arc $TX$ on $(ATX)$, then $K' \in XK$. This means that the spiral similarity at $A$ taking $\triangle ATX$ to $\triangle ABL$ will also take $K'$ to $K$. Now, we can see that $$\angle TPQ-\angle K'TQ = (\angle XTQ-\angle XTK')-(\angle BTP- \angle TXP) = \angle XTQ-\angle BTP$$so it suffices to prove that $\angle TPQ = \angle K'TQ$, in particular, that $(TQP)$ is tangent to $TK'$ at $T$. $\textbf{Claim:}$ $K'T^2 = K'Q \cdot KP$ $\textbf{Proof)}$ The idea is to further use the spiral similarity at $A$. Let $P', Q'$ be the images of $P,Q$ under this spiral similarity. As $XK' \to LK$, we have that $P',Q'$ lie on $KL$. We have that $KB^2 = KP \cdot KQ$ and wish to prove that $KB^2 = KP' \cdot KQ'$ because $Q \to Q', P \to P', T \to B$ under this spiral similarity. In particular, $QPQ'P'$ has to be a cyclic quadrilateral with Miquel point $A$, which is equivalent to the concurrence of $AK$, $QQ', PP'$ on $(APQ)$ which can be proven by firstly showing that $BL, AK, (APQ)$ concur at a point $T$ (because $\angle LBK = \angle BAK$ by the tangent chord theorem) and then introducing $Q*, P*$ as the intersections of $QT$ and $PT$ with $KL$ and showing that $Q* = Q', P* = P'$ (the interested reader can check this carefully, but this part is not really the point of the problem; the creative part is to further use the spiral similarity by considering where $P$ and $Q$ go and noticing that we will get a cyclic quadrilteral - rest being just the usual Miquel point config). $\blacksquare$ The claim clearly implies the tangency of $(TQP)$ to $TK'$ at $T$. $\blacksquare$ $\blacksquare$

Invert from $B$ with radius $BK$. $A^*T^*\parallel BL^*, \ \overline{PQA}\parallel BK$ and $A,K,L,X$ are collinear. Also $B,X^*,P^*,Q^*,K$ are cyclic. $P^*Q^*\cap BL^*=Y$ then $YA^*BK$ is an isosceles trapezoid. Since $P^*Q^*\parallel BK$ and $B,K,P^*,Q^*$ are cyclic, $P^*Q^*BK$ is an isosceles trapezoid. $\angle BTQ=\angle T^*Q^*B$ and $\angle PTX=180-\angle BTP=180-\angle T^*P^*B$ hence we want to show that $\angle T^*Q^*B+\angle T^*P^*B=180$. New Problem Statement: Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$ and $AB<CD$. Take $P,Q$ on $CD$ such that $ABPQ$ is an isosceles trapezoid. $F$ is the second intersection of $BD$ with $(ABPQ)$. $M$ is on $AF$ such that $DM\parallel AC$. Then, $\angle MPA+\angle MQA=180$. Proof: Let $AF\cap CD=N$. \[\angle FDN+\angle PQF=\angle DBA+\angle PAF=\angle FPA+\angle PAF=\angle PFN\]Hence $(FDN)$ and $(FPQ)$ are tangent to each other. Let the common tangent at $F$ intersect $CD$ at $S$. Since $DM\parallel AC,$ we have $\angle NDM=\angle CDM=\angle DCA=\angle BDC=\angle FDN$. We have $\angle SFN=\angle FDN=\angle NDM=\angle SDM$ thus, $S,F,D,M$ are cyclic. $\angle SFM=\angle SDM=\angle FDS=\angle FMS$ yields $SF=SM$. \[SM^2=SF^2=SP.SQ\]Hence $\angle SMP=\angle MQP$ \[\angle MPD+\angle MQD=\angle MPS+\angle SMP=\angle MSD=\angle MFD=\angle AFB=\angle AQB\]\[\angle MPA+\angle MQA=\angle MPD+\angle DPA+\angle MQD+\angle PQA=\angle AQB+(180-\angle APQ)+\angle PQA=180\]As desired.$\blacksquare$