Fact. If $x$, $y$, $z$ are points on the complex plane, then the circumcenter of $\triangle xyz$ is the point
$$ i \frac{x(z\overline{z} - y\overline{y})+ y(x\overline{x}-z\overline{z})+z(y\overline{y}-x\overline{x})}{x(\overline{z} - \overline{y})+ y(\overline{x}-\overline{z})+z(\overline{y}-\overline{x})}$$
where "$\overline{x}$" indicates the complex conjugate of $x$.
Let us consider $\triangle abc$, defining $d$, $e$, $f$ on sides $bc$, $ca$, $ab$, respectively:
$$d := b + \alpha(c-b) \qquad e := c+\beta(a-c) \qquad f := a+\gamma(b-a)$$
for real scalars $\alpha, \beta, \gamma$.
Let $p$, $q$, $r$ be the respective circumcenters of $\triangle aef$, $\triangle dbf$, $\triangle dec$; for simplicity, we take $a$, $b$, $c$ on the unit circle so that $\overline{a}=1/a$, etc. Then,
$$\begin{align}
p &= i\;\frac{c(b-a)+ \beta b ( a-c) + \gamma c( a - b )}{b-c} \\[6pt]
q &= i\;\frac{a(c-b)+ \gamma c ( b-a) + \alpha a( b - c )}{c-a} \\[6pt]
r &= i\;\frac{b(a-c)+ \alpha a ( c-b) + \beta b( c - a )}{a-b}
\end{align}$$
We deduce
$$\frac{p-q}{a-b} = \frac{q-r}{b-c} = \frac{r-p}{c-a} =: u$$
which, by taking the modulus, proves $\triangle abc \sim \triangle pqr$. A little algebra reveals that
$$u + \overline{u} = 1$$
Now ... A point, $m$, (which may or may not be the Miquel point) is the center of spiral similarity from $\triangle abc$ and $\triangle pqr$, if and only if
$$\frac{a-m}{p-m} = \frac{b-m}{q-m} = \frac{c-m}{r-m}$$
(This characterization is what inspired me to approach the problem in the complex plane.) Solving for $m$ via the first equality gives
$$m = \frac{aq-bp}{(a-b)-(p-q)} = \frac{r - c u}{1-u}$$
whence
$$m-r = \frac{u(r-c)}{1-u} = \frac{u}{\overline{u}}(r-c) \qquad\qquad
\overline{m}-\overline{r} = \frac{\overline{u}}{u}(\overline{r}-\overline{c})$$
so that
$$|m-r|^2 = (m-r)(\overline{m}-\overline{r}) = (r-c)(\overline{r}-\overline{c}) = |c-r|^2$$
indicating that $c$ and $m$ are equidistant from $r$: thus, $m$ is on the circumcircle of $\triangle dec$. By symmetry, it is also on the circumcircles of $\triangle aef$ and $\triangle dbf$; the center of spiral similarity is in fact the Miquel point.
Here is the picture of the story, so that the community can see where the composer of the problem placed almost all letters of the alphabet.
The picture also reveals some points that the problem missed to introduce, although this would have been a good chance to use the alphabet, so the composer really wanted to make a puzzle out of it.
Psychological digression. Please skip, if the next few sentences feel annoying. This insertion is only a psychological hint and survivor guide for potential Mathematics Olympiad problems in the same style.
I have often seen such problems in my youth, the "composer" considers a given situation with easily introduced points, that have easy properties and are introduced in an easy order. Form the easy properties (for instance intersection of lines) one obtains more complicated properties (the intersection of lines lies on one circle). Then one reverts the order of introduction of points, and each point is introduced in the most complicated possible manner. Now one can proceed in two ways. One can "accept the challenge", and use the introduced properties with all difficulties that may occur giving the "sincere proof", or one ignores the statement, for "difficult points" $X,Y,Z\dots$ in the problem one introduces (apparently with no connection to the problem) own points $X', y', Z', \dots$ possibly in other order, shows a lot of simple facts with $X', y', Z', \dots$, then finally shows (possibly in other order) that $X=X'$, then $Y=Y'$, then $Z=Z'$, then... (by uniqueness of the construction), and suddenly the problem is solved without pain. This is the reverse engineering attempt to solve, maybe mirroring exactly the way the problem was "composed".
Note that in the given case, the points $B,C$ have no role, so we will delete them from the picture. Such "beauty errors" are considered in chess composition, well, a displaced parallel maybe, as "true errors", and the composer is urged to restate by giving only the essential. But we are here not in a beauty contest, and even more, it is an important step in the mathematical thinking to restrict to the essential, so let us do this.
We will solve the problem by introducing only some of the letters, and showing that the two circles $(TAN)$ and $(LAS)$ are tangent in $A$, the common tangent being... $AD$. Doing this feels already like solving only half of the problem. We use the symmetry / the similarity of properties of the points involved in the construction.
Let us restate, and after each construction we make some simple observations outside the statement, so that finally the statement comes with solution. We show first the picture:
Statement (with immediate comments):
Let $\Delta DEF$ be a triangle. For reasons that will be transparent soon, we may like to use $E^*=F$ as an alternative notation.
Let $(I)$ be the circle through $D,E,F$, centered at $I$, so $ID=IE=IF$.
The tangents in $E$ and $E^*=F$ to $(I)$ intersect in $A$.
In particular, $AI$ is diameter, the angles in $E,E^*$ in the two triangles with hypotenuse $AI$ are right angles, and $\hat D:= \widehat{E^*DE}=\frac 12\widehat{E^*IE} =\frac 12(\pi-\widehat {EAE^*})=\frac 12\pi -\frac 12\hat A=\widehat{EIA}=\widehat{AIE^*}$.
We draw the circle $(AEIE^*)$. The lines $DE$ and $DE^*$ intersect this circle for the second time in points denoted by $U$ and respectively $U^*$.
Then $\widehat{EUA}=\widehat{EIA}=\widehat{EDE^*}$, so $AU\| E^*U^*D$, and similarly $AU^*\|EUD$. So $AUDU^*$ is a parallelogram. In particular $AE=AE^*=UU^*$, because the six points lie on the circle, and correspond to equal angles, $\widehat{AIE^*}=\widehat{AIE}=(\hat D=\widehat{U^*DU}=)\widehat{U^*AU}$.
Let $X$ be the second point on $AD$ which lies on the circle $DEE^*$.
We mark the angle $\widehat{E^*AD}$ in blue so that we find this angle at some other places: $\widehat{E^*AD}=\widehat{ADE}=\widehat{XDE}=\widehat{XE^*E}$.
We draw the line $E^*X$, it intersects $AU$ in $S$. On the other side:
We draw the line $EX$, it intersects $AU^*$ in $S^*$.
Then $SXS^*A$ is cyclic, because the sum of the opposite angles in $A$ and $X$ is the same as the sum of the opposite angles in $D$ and $X$ of the cyclic $XEDE^*$. In particular we can add one more blue angle to the list, since $\widehat{S^*AX}=\widehat{S^*SX}$, so we get one more pair of parallel lines, $SS^*\|EE^*$. The triangles $\Delta ASS^*$ and $\Delta DUU^*$ are thus in perspective, and the point of perspective is $X=AD\cap ES^*\cap E^*S$.
The angle in $D$ was marked in green. Let us show that a pair of opposite angles in $X$ can be marked in green, too. We have $\widehat{SXE}=\widehat{XE^*E}+\widehat{XEE^*}=\widehat{XDE}+\widehat{E^*DX}=\widehat{E^*DE}=\hat D$.
The line $E^*XS$ further intersects the circle $(E^*AEI)$ in a point denoted by $M$.
Then we can add one more blue angle to the list, because
$\widehat{MUE}=\widehat{ME^*E}$, and from $\widehat{MUE}=\widehat{ADE}$ we deduce $MU\| AD$.
On the other side, the line $EXS$ further intersects the circle $(EAE^*I)$ in a point denoted by $M^*$.
Similarly we get the parallelism $M^*U^*\|AD$, so $AD$, $MU$, $M^*U^*$ are parallel.
Note that the angles in $M, M^*$ in $AMXM^*$ are equal and "green", because they correspond to equal chords $AE^*=AE$. The exterior angles in $X$ are also "green", so $AMXM^*$ is a parallelogram. So we may display:
$AM\|EXS^*M^*$ and $AM^*\|E^*XSM$.
Let $L,L^*$ be the intersections of $SS^*$ with $AM$, respectively with $AM^*$.
Then the triangles $\Delta ALL^*$ and $\Delta XEE^*$ are similar and have pairs of sides respectively parallel.
They are in perspective, so $AX,LE,L^*E^*$ intersect in a point. We have only to show that this point is $D$. Let us see.
We have one more blue angle in the list, $\widehat{LL^*A}=\widehat{SL^*A}
=\widehat{}L^*SE^*$, since $LL^*\|EE^*$.
In particular, $\widehat{LL^*A}=\widehat{LDA}$, so $ALDL^*$ cyclic, exactly
as $XEDE^*$. The point $D$ is common for the two cycles, and lies on $AX$, so
it is the center of the perspective of $\Delta ALL^*$ and $XEE^*$. So
$$
AX\cap LE\cap L^*E^*=D\ .
$$
Let $Q,Q^*$ be the midpoints of the segments $AE$, respectively $AE^*$.
Then they are the intersection of the diagonals in the parallelograms
$ALES^*$ and $AL^*E^*S$. So the six points $S,S^*;L, L^*;Q, Q^*$ are on a line. But we do not need $Q,Q^*$ in the following.
Claim: The circles $(ASL)$ and $(AS^*L^*)$ are tangent in $A$, and the common tangent is the line $AXD$.
The proof is now simple, note that the equality of blue angles $\widehat{AL^*S^*}=S^*AD$ implies that $AD$ is tangent in $A$ to $\Delta AL^*S^*$. Same game on the other side.
$\square$
It remains to check that the complicated constructions in the OP lead to the same points.
Best Answer
Hint: We recognize that the circumcircles are hard to work with directly. Instead, let's consider the points $O^*, A^*, B^*, C^*$, which are obtained by expansion from point $P$ by a factor of 2.
Judging from the phrasing of your question, I think you have shown that the 4 circles are concurrent. If not, here is an approach: