This is a solution for part 1. I haven't thought about part 2 yet.
The following text may not be very rigorous. To be more specific, the solution may rely on the picture a bit. For example, it may implicitly use the fact that points $P$, $E$ and $F$ all lie on the same side of line $BC$, simply because it looks that way on the picture. I'm afraid that if I make the solution any more rigorous, it will become absolutely incomprehensible.
Step 1. Build $P'$ and several other new points.
We will forget about point $P$ and the bigger circle $\odot O_1$ for a while, and focus on the triangle $\triangle ABC$ and the smaller circle $\odot O_2$. Consider the circumcircle of $\triangle EBD$. It intersects $\odot O_2$ at point $E$. There must be a second point where these two circles intersect. Let us denote that point by $P'$.
Clearly, $E,\,B,\,D,\,P'$ are concyclic. Our plan is simple: to prove that $P=P'$. To do this, we'll construct several more lines and points. Let us denote by $Q$ the second point where line $P'D$ intersects $\odot O_2$. Let $B'$ and $C'$ be the points of intersection of $O_2$ with $P'B$ and $P'C$ respectively. You can see all these on the figure:
Step 2. Explore the properties of the newly constructed objects.
The objective of this step is to prove that $BC||B'C'$. We will do this by some simple angle chasing.
First of all, since $EBDP'$ is an inscribed quadrilateral,
$$
\angle EP'Q=\angle ABD = 1/2 \angle ABC.
$$
It is quite easy to see that
$$
\angle EP'F = 90^{\circ} - 1/2 \angle BAC = 1/2(\angle ABC + \angle ACB).
$$
From this we have
$$
\angle FP'D = 1/2 \angle ACB = \angle ACD,
$$
so $FCDP'$ is also an inscribed quadrilateral.
So, we have two inscribed quadrilaterals $EBDP'$ and $FCDP'$. From the first one we know that $\angle BED = \angle BP'D$ and from the second that $\angle CFD = \angle CP'D$. It is obvious that $\angle BED = \angle CFD$ (because $\triangle AED = \triangle AFD$), therefore $\angle BP'D = \angle CP'D$. To put it in different letters, $\angle B'P'Q = \angle C'P'Q$. This in turn means that $\angle B'O_2Q = \angle C'O_2Q$, which means that $B'C' \perp O_2Q$. Now to prove that $B'C'||BC$, it only remains to prove that $O_2Q \perp BC$.
But this is easy. We already know that $\angle EP'Q = 1/2 \angle ABC$. Also, $\angle EP'Q = 1/2 \angle EO_2Q$. So $\angle EO_2Q = \angle ABC$. Since $O_2E \perp BA$, it means that $O_2Q \perp BC$, qed.
Step 3. Prove that $P=P'$.
Alright, we now know that $BC||B'C'$. From here it will be quite easy to prove that $P=P'$.
Since $BC||B'C'$, there exists a homothety $f$ with center $P'$ that sends line $B'C'$ to line $BC$. Clearly, $f(B')=B$ and $f(C')=C$. Let us denote $\odot O'_1 = f(\odot O_2)$. Since $\odot O'_1$ is the image of $\odot O_2$ under a homothety with center $P'$, circles $\odot O'_1$ and $\odot O_2$ are tangent at $P'$. Also, since $B'$ and $C'$ lie on $\odot O_2$, $B$ and $C$ lie on $\odot O'_1$.
Let us revise what we have established. We have two circles $\odot O_1$ and $\odot O'_1$. Both of them contain points $B$ and $C$ and both of them are tangent to $\odot O_2$ at points $P$ and $P'$ respectively. But from this it is already clear that $\odot O_1$ and $\odot O'_1$ are the same, and so are points $P$ and $P'$. Done!
After some symbol-bashing in Mathematica, here are barycentric coordinates of key points:
$$\begin{align}
A' &= (0 : a + b - c : a - b + c) \\[0.75em]
A'' := \tfrac12(B'+C') &=
\left(a +\frac{(a-b+c)(a+b-c)}{-a+b+c} : b : c\right) \\[0.75em]
A''' &= (2 a + b + c : b : c ) \\[0.75em]
O_A &= (2 a^3 : a^3 - b^3 + c^3 + a^2 b - a^2 c - 3 a b^2 - a c^2 - b^2 c + b c^2 - 4 a b c \\
&\phantom{(2 a^3\;}\quad : a^3 + b^3 - c^3 - a^2 b + a^2 c - a b^2 - 3 a c^2 + b^2 c -
b c^2 - 4 a b c)
\end{align}$$
The coordinates of the points where, say, $\bigcirc O_A$ (the one passing through incenter $X_1$) meets the side-lines of the circle are a little messy, so I'll omit them. Nevertheless, the six prescribed points are indeed cyclic, and the equation of their common circle has this barycentric form (in $u:v:w$ coordinates):
$$\begin{align}
0 &= u^2 b c (-a + b + c) \cos A + v^2 c a (a - b + c) \cos B + w^2 a b (a + b - c) \cos C \\[0.75em]
&\quad+ v w (-a^3 + b^3 + c^3 - 2 a^2 b - 2 a^2 c - b^2 c - b c^2) \\[0.75em]
&\quad+ w u (-b^3 + c^3 + a^3 - 2 b^2 c - 2 b^2 a - c^2 a - c a^2) \\[0.75em]
&\quad+ u v (-c^3 + a^3 + b^3 - 2 c^2 a - 2 c^2 b - a^2 b - a b^2)
\end{align}$$
Converting to trilinear form (in $\alpha:\beta:\gamma$ coordinates), the equation can be manipulated into this:
$$(\lambda \alpha + \mu\beta+\nu\gamma)(a\alpha+b\beta+c\gamma)+\kappa(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$
where
$$\begin{align}
\lambda &:= (-a + b + c)\cos A \\
\mu &:= (\phantom{-}a-b+c)\cos B\\
\nu &:= (\phantom{-}a+b-c)\cos C\\
\kappa &:= -2(a+b+c)
\end{align}$$
As $\lambda:\mu:\nu$ are the trilinear coordinates for Kimberling center $X(219)$ ("$X(8)$-Ceva Conjugate of $X(55)$"), the circle is the Central Circle of that point. This answers the "alternatively" aspect of OP's question.
I don't know what Kimberling centers may lie on the circle. I don't even know how one would go about searching for them. Someone with more fluency in triangle-center lore may have some insights.
Incidentally, for the $\bigcirc O_A$ passing through $B''$, $B'''$, $C''$, $C'''$, etc, the corresponding six-point circle has trilinear form
$$(\lambda'\alpha+\mu'\beta+\nu'\gamma)(a\alpha+b\beta+c\gamma)+\kappa'(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$
where
$$\begin{align}
\lambda' &:= a (-a + b + c)^2 \\
\mu' &:= b (\phantom{-}a - b + c)^2 \\
\nu' &:= c (\phantom{-}a + b - c)^2 \\
\kappa' &:= -8 a b c
\end{align}$$
This circle is therefore the Central Circle of Kimberling's $X(220)$ ("$X(9)$-Ceva conjugate of $X(55)$").
I have not investigated whether there is a general connection between OP's construction, Ceva-conjugates, and/or $X(55)$.
Best Answer
The key is to realize that this three-dimensional problem is really just a folded two-dimensional problem. Unfold the figure along hinge $AD$ to get a nice, flat $\triangle ABC$.
Define $\beta := \angle BAD$ and $\gamma := \angle CAD$. As a minor convenience, we'll assume $\angle A$ is non-obtuse (the obtuse case is largely the same), and define $\alpha := 90^\circ-\beta-\gamma$.
The condition $\angle YCD=\angle DAB\;(=\beta)$ implies that $\overline{CY}\perp\overline{AB}$; this makes $Y$ the orthocenter of our flattened triangle, which in turn implies $\overline{BY}\perp\overline{AC}$ and $\angle YBD=\angle DAC\;(=\gamma)$. Moreover, $\angle ABY=\angle ACY=\alpha$.
Define $X'$ as the reflection of $X$ in $\overline{BC}$. The circle conditions imply that $BMNXX'$ and $CPQXX'$ are each cyclic. The fact that the circles have common chord $\overline{XX'}$ whose extension contains $A$ implies that $A$ has a common power with respect to both circles; likewise for $Y$. Thus, we can write $$\begin{align} |AB||AM|=|AC||AP| &\quad\to\quad \frac{|AB|}{|AC|}=\frac{|AP|}{|AM|} \quad\to\quad \triangle ABC \sim \triangle APM \tag1\\[6pt] |YB||YN|=|YC||YQ| &\quad\to\quad \frac{|YB|}{|YC|}=\frac{|YQ|}{|YN|} \quad\to\quad \triangle YBC \sim \triangle YQN \tag2 \end{align}$$
Matching-up corresponding angles in the similar triangles, we have $$\begin{align} \angle AMP &=\angle ACB=\alpha+\beta \tag3\\ \angle APM &=\angle ABC=\alpha+\gamma \tag4 \\[8pt] \angle YNQ &=\angle YCB=\beta \quad\to\quad \angle BMN = \beta \quad\to\quad \angle AMN=\alpha+\beta \tag5 \\ \angle YQN &=\angle YBC=\gamma \quad\to\quad \angle CQP \;\,= \gamma \quad\to\quad \angle APQ \;\,= \alpha+\gamma \tag6 \end{align}$$
Since $\angle AMP = \angle AMN$, we have that $M$, $N$, $P$ are collinear; likewise for $M$, $Q$, $P$. Let the line through these four points meet $\overline{AD}$ at $Z$; since $Z$ is on the extended common chord $\overline{XX'}$, we have $$|ZM||ZN| = |ZP||ZQ| \tag7$$
Finally, re-fold the figure along $\overline{AD}$ to restore the three-dimensional tetrahedron. The collinearity of $M$, $N$, $P$, $Q$, $Z$ becomes coplanarity. Moreover, taking the left-hand side of $(7)$ as computing the power of $Z$ with respect to the circumcircle of $\triangle MNP$, the equality with the right-hand side implies that $Q$ must lie on that circle. Done!