1. **Claim 1: Let $G$ be the intersection of the angle bisector of $\angle DEA$ and $AD$, then $LN \parallel EG$.

Proof:**
- Consider the Miquel Point of $ABCD$. By the properties of the Miquel Point, we have:
$$\frac{BK}{KA} = \frac{BC}{DA} = \frac{PB}{PA}$$
Hence, $PK$ is the angle bisector of $\angle BPA$, and the cyclic variants hold by symmetry.
- Therefore, $P$ is also the center of spiral similarity sending $MK$ to $DA$, hence $P, E, M, K$ are concyclic. Similarly, $P, L, N, F$ are concyclic.
- Now, we can chase the angles:
$$\angle LNG = \angle FPL = \angle FPC + \frac{1}{2} \angle CPL = \angle EDA + \frac{1}{2} \angle CEB = \angle EGA$$
Thus, $LN \parallel EG$. $\blacksquare$

2. **Claim 2: $K, L, M, N$ concyclic implies $ABCD$ is a circumscribed quadrilateral.

Proof:**
- We have:
$$\angle MNL = \angle DNL - \angle DNM = \angle DGE - \angle DNM$$
and
$$\angle MKL = \angle BKM - \angle BKL = \angle BHF - \angle BKL$$
Since they are equal:
$$\angle BKL - \angle DNM = \angle BHF - \angle DGE = \frac{1}{2} \angle BFD + \angle BAD - \frac{1}{2} \angle DEA - \angle BAD = \frac{1}{2} (\angle CDA - \angle CBA) \quad (1)$$
Rearranging gives:
$$\angle DNM + \frac{1}{2} \angle MDN = \angle BKL + \frac{1}{2} \angle LBK$$
Let $B_1$ be the intersection of the angle bisector of $\angle LBK$ and $LK$. Define $D_1$ similarly. Then from (1) we have:
$$\angle MD_1D = \angle BB_1L$$
On the other hand, we have:
$$\frac{BL}{BK} = \frac{NA}{AK} = \frac{DN}{DM} = \frac{CM}{CL} = k$$
for some $k$. Suppose on the contrary that $k > 1$, then $DN > DM$ and $BL > BK$, hence $\angle MD_1D > 90^\circ > \angle BB_1L$, contradiction. Similarly, $k < 1$ is impossible. This implies $k = 1$ so $DN = DM$, $CM = CL$, $NA = NK$, and $BL = BK$ so we are done. $\blacksquare$

3. **Claim 3: If $ABCD$ is circumscribed then the incircle of $\triangle CEF$ and $\triangle AEF$.

Proof:**
- This is just length chasing. It suffices to show:
$$AE - AF = EC - EF$$
Indeed, $AE - AF = EB + BA - DA - DF = EB + BC - CD - DF$, hence it suffices to show:
$$EB + BC - EC = CD + DF - CF \quad (2)$$
Notice that $L$ is the touching point of the $E$-excircle and $BC$, hence the left-hand side of (2) equals $CL$ while the right-hand side of (2) equals $CM$. Obviously, they are equal so we are done. $\blacksquare$

4. Final Step:

- Now let $Z$ be the intersection of $UV$ and $EF$. Then $(E, F; J, Z) = -1$. Therefore, $T, S, Z$ are collinear. As a result:
$$ZT \times ZS = ZJ^2 = ZU \times ZV$$
hence $S, T, U, V$ are concyclic as desired. $\blacksquare$