Sketch: Draw the circumshpere of ABCD and plane PA0B0C0D0. Let l cross shpere ABCD at X and Y. Use monge theorum and we can get A0 B0 C0 D0 P X Y are concyclic.

I simply don't have even the basic ideas of projective transformations apart from very little exposure given in EGMO. Can someone please tell me if this works? We first perform an inversion with center $P$ any arbitrary radius. Let $A^*,B^*,C^*,D^*$ denote the inverses of $A,B,C,D$. So, the sphere $PBCD$ gets mapped to the plane $B^*C^*D^*$. Thus $A_1=PA_0\cap\operatorname{sphere}(PBCD)\mapsto PA_0\cap\operatorname{plane}(B^*C^*D^*)=A_1^*$ (say). Define $B_1^*,C_1^*,D_1^*$ similarly. Thus the condition $P,A_1,B_1,C_1,D_1$ are cyclic changes to proving $\overline{A_1^*-B_1^*-C_1^*-D_1^*}$ are collinear. Now we simply drop the condition that $A^*,B^*,C^*,D^*$ are the inverses of $A,B,C,D$, i.e., we put $A^*,B^*,C^*,D^*$ as any arbitrary point on $PA,PB,PC,PD$! Thus our problems gets the following form (Yes, this actually works, I checked in 3D GeoGebra lol). $\textbf{PROBLEM: }$Given a tetrahedron $ABCD$ and a line $\ell$ intersecting the planes $ABC,BCD,CDA,DAB$ at points $D_0,A_0,B_0,C_0$. Let $A^*,B^*,C^*,D^*$ be arbitrary points on $PA,PB,PC,PD$ respectively. Let $A_1^*=PA_0\cap\operatorname{plane}(B^*C^*D^*)$. Define $B_1^*,C_1^*,D_1^*$ similarly. Then $\overline{A_1^*-B_1^*-C_1^*-D_1^*}$ are collinear. We firstly prove that $\overline{A_1^*-B_1^*-C_1^*}$ are collinear. We can derive the others in a similar manner thus concluding $\overline{A_1^*-B_1^*-C_1^*-D_1^*}$ are collinear. Firstly, we take a homothety centered at $P$ that maps $D^*\mapsto D$. Note that all collinearities are still preserved. Thus we can give one final restatement. $\textbf{PROBLEM: }$Given two triangles $ABC$ and $A'B'C'$ in the 3D space such that they are perspective to each other at some point $P$. $D$ be another distinct point in the space. A line $\ell$ intersects the planes $DBC,DCA,DAB$ at $A_0,B_0,C_0$. Now define $A_1=PA_0\cap\operatorname{plane}(DB'C')$. Similarly define $B_1,C_1$. Then $\overline{A_1-B_1-C_1}$ are collinear. We note that our current problem is purely projective! So we take a projective transformation by projecting from $D$ onto the plane of $ABC$. We have the following problem after the transformation. $\textbf{PROBLEM: }$Let $ABC$ and $A'B'C'$ be two triangles in the projective plane that are perspective to each other at $P$. A line $\ell$ intersects sides $BC,CA,AB$ at $A_0,B_0,C_0$ respectively. Now $A_1=PA_0\cap B'C'$. Define $B_1,C_1$ similarly. Then $\overline{A_1-B_1-C_1}$ are collinear. $\textbf{SOLUTION: }$Firstly, let $X=BC\cap B'C'$, $Y=CA\cap C'A'$ and $Z=AB\cap A'B'$. By Desargues' Theorem, we have that $\overline{X-Y-Z}$ are collinear. Also we had that $\overline{A_0-B_0-C_0}$ are collinear too. So by Menelaus, we have that, \begin{align*} &\dfrac{AC_0}{C_0B}\cdot\dfrac{BA_0}{A_0C}\cdot\dfrac{CB_0}{B_0A}=-1\\ &\dfrac{AZ}{ZB}\cdot\dfrac{BX}{XC}\cdot\dfrac{CY}{YA}=-1\\ &\dfrac{A'Z}{ZB'}\cdot\dfrac{B'X}{XC'}\cdot\dfrac{C'Y}{YA'}=-1. \end{align*} Now $(A,B;C_0,Z)\overset{P}{=}(A',B';C_1,Z)$. Moreover, we work with directed lengths from now. \begin{align*} &\implies\dfrac{C_0A}{C_0B}\div\dfrac{ZA}{ZB}=\dfrac{C_1A'}{C_1B'}\div\dfrac{ZA'}{ZB'}\\ &\implies\dfrac{C_1A'}{C_1B'}=\dfrac{C_0A}{C_0B}\cdot\dfrac{ZB}{ZA}\cdot\dfrac{ZA'}{ZB'}\\ &\implies\dfrac{A'C_1}{C_1B'}=\dfrac{AC_0}{C_0B}\cdot\dfrac{ZB}{AZ}\cdot\dfrac{A'Z}{ZB'}. \end{align*} Similarly, we also get, \begin{align*} \dfrac{C'B_1}{B_1A'}&=\dfrac{CB_0}{B_0A}\cdot\dfrac{YA}{CY}\cdot\dfrac{C'Y}{YA'}\\ \dfrac{B'A_1}{A_1C'}&=\dfrac{BA_0}{A_0C}\cdot\dfrac{XC}{BX}\cdot\dfrac{B'X}{XC'}. \end{align*} So we now have, \begin{align*} \dfrac{A'C_1}{C_1B'}\cdot\dfrac{C'B_1}{B_1A'}\cdot\dfrac{B'A_1}{A_1C'}&=\left(\dfrac{AC_0}{C_0B}\cdot\dfrac{ZB}{AZ}\cdot\dfrac{A'Z}{ZB'}\right)\cdot\left(\dfrac{CB_0}{B_0A}\cdot\dfrac{YA}{CY}\cdot\dfrac{C'Y}{YA'}\right)\cdot\left(\dfrac{BA_0}{A_0C}\cdot\dfrac{XC}{BX}\cdot\dfrac{B'X}{XC'}\right)\\ &=\left(\dfrac{AC_0}{C_0B}\cdot\dfrac{BA_0}{A_0C}\cdot\dfrac{CB_0}{B_0A}\right)\cdot\left(\dfrac{ZB}{AZ}\cdot\dfrac{XC}{BX}\cdot\dfrac{YA}{CY}\right)\cdot\left(\dfrac{A'Z}{ZB'}\cdot\dfrac{B'X}{XC'}\cdot\dfrac{C'Y}{YA'}\right)\\ &=-1\cdot-1\cdot-1\\ &=-1. \end{align*}And we are done by the converse of Menelaus.

I am very happy that this interesting 3D geometry problem of mine was selected by the Organizing Committee. The official solution is published here. I am sending the solution I submitted as follows: Let $\mathcal{S}$ be the circumsphere of the tetrahedron $ABCD$. Let $\mathcal{C}$ be circumcircle of the triangle $PA_1B_1$. We easily seen plane $(BCD)$ is the radical plane of $\mathcal{S}$ and the circumsphere of tetrahedron $PBCD$, also $A_0$ lies on plane $(BCD)$, thus power of $A_0$ with respect to $\mathcal{S}$ equal to power of $A_0$ with respect to circumsphere of tetrahedron $PBCD$, which is the product $\overline{A_0P}\cdot\overline{A_0A_1}$ (because line $PA_0$ meets circumsphere of tetrahedron $PBCD$ again at $A_1$). Also the product $\overline{A_0P}\cdot\overline{A_0A_1}$ is the power of $A_0$ with respect to circle $\mathcal{C}$ (circumcircle of the triangle $PA_1B_1$). Thus, $A_0$ lies on the radical axis of $\mathcal{S}$ and $\mathcal{C}$. Prove similarly, $B_0$ lies on the radical axis of $\mathcal{S}$ and $\mathcal{C}$. Thus, line $\Delta$ is the radical axis of $\mathcal{S}$ and $\mathcal{C}$. Therefore $C_0$ must have the same power with respect to $\mathcal{S}$ and $\mathcal{C}$. Since $C_0$ lies on the plane $(DAB)$ (which is radical plane of $\mathcal{S}$ and circumcircle of tetrahedron $PDAB$), power of $C_0$ with respect to $\mathcal{S}$ is equal to power of $C_0$ with respect to circumsphere of tetrahedron $PDAB$. From these, line $PC_0$ must be the radical axis of circle $\mathcal{C}$ and circumsphere of tetrahedron $PDAB$, but line $PC_0$ meets circumsphere of tetrahedron $PDAB$ at $C_1$, this means $C_1$ lies on the circle $\mathcal{C}$. Prove similarly, point $D_1$ lies on the circle $\mathcal{C}$. We get fives concyclic points $A_1$, $B_1$, $C_1$, $D_1$, and $P$. P/s: If we use polar inversion $P$, then this can be considered as another solution to this problem (see simple proof of kiyoras_2001).