Excellent angle chasing exercise. We do something similar to what I did on APMO 2024 p5 (point spam lol), in order to connect everything, so define $(APD) \cap (BPC)=Q$, let $(APD)$ hit $AB,CD$ at $G,J$ respectively and $(BPC)$ hit $AB,CD$ respectively at $H,I$, also let $DD' \cap C'B=E$ and $AD' \cap CC'=F$ and finally let $AC' \cap CD'=K$. Fitst from double miquel theorem on triangles $\triangle BDE, \triangle ACF$ we get that $FD'QC'E$ is cyclic, so from a LOT of reim's we get that $F,J,Q,H$ and $G,Q,I,E$ are colinear quadruples, now to finish note that \[ \angle QC'D'=\angle D'FQ=\angle AD'Q-\angle D'AJ=\angle ADQ-\angle ADP=\angle PDQ=\angle PAQ \]And similarily we can obtain that $\angle QD'C'=\angle QCA$ which means that $Q$ is miquel of $AC'D'C$ therefore $D'KQC'$ and $AKQC$ are cyclic so to finish: \[\angle XPC=\angle AD'K=\angle KC'F=\angle YPC \implies X,Y,P \; \text{colinear!} \]Therefore we are done .

Let $DD'\cap CC'=S$ and $(PAD)\cap (PBC)=Q$. Claim: $\measuredangle AC'B+\measuredangle DC'C=180$. Proof: Let $E$ be a point on the ray $C'P$ with $\frac{PE}{PC'}=\frac{PA}{PC}=\frac{PB}{PD}$. Note that $PAEB\sim PCC'D$. \[\measuredangle EAB=\measuredangle C'CD=\measuredangle PCB=\measuredangle PC'B=\measuredangle EC'B\]Hence $E,C',A,B$ are concyclic. This implies $180=\measuredangle BEA+\measuredangle AC'B=\measuredangle DC'C+\measuredangle AC'B$ which completes the proof of the claim.$\square$ Claim: $P,Q,S$ are collinear. Proof: Let $R$ be the image of $S$ under the homothety centered at $P$ sending $CD$ to $AB$. We can see that $R,P,S$ are collinear. \[\measuredangle AQB=\measuredangle AQP+\measuredangle PAB=\measuredangle ADB+\measuredangle ACB=\measuredangle CDS+\measuredangle SCD=180-\measuredangle DSC=180-\measuredangle BRA\]Thus, $Q,R,A,B$ are concyclic. \[\measuredangle AQP=\measuredangle ADB=\measuredangle CDS=\measuredangle ABR=\measuredangle AQR\]Which proves the collinearity of $P,Q,R,S$.$\square$ We conclude that $SD'.SD=SP.SQ=SC'.SC$ hence $C,D,C',D'$ are concyclic. The problem wants us to prove $\measuredangle AD'C+\measuredangle AC'C=180$. \[\measuredangle AD'C=\measuredangle AD'S-\measuredangle CD'S=(180-\measuredangle DD'A)-\measuredangle SC'D=\measuredangle DC'C-\measuredangle DPA\]\[180-\measuredangle AD'C=180+\measuredangle DPA-\measuredangle DC'C=\measuredangle DQA+\measuredangle AC'B=\measuredangle BQC+\measuredangle AC'B\]\[\measuredangle BQC+\measuredangle AC'B=\measuredangle AC'B+\measuredangle BC'C=\measuredangle AC'C\]Thus, $\measuredangle AC'C+\measuredangle AD'C=180$ as desired.$\blacksquare$

Let $G$ be the center of spiral similarity taking $ABC$ to $XYZ$ and $K = BY \cap CZ$. Note that $G \in (KYZ), (KBC), (FAXE)$ by basic spiral similarity config. Let $J = KX \cap (KBC)$ and $Q = KX \cap (KYZ)$. Note that the spiral similarity defined previously takes $J$ to $Q$. $\angle GJQ$ is the rotation angle of this spiral similarity as is $\angle GAX$ which means that $J \in (GAX)$ as desired.