Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

$\angle AA'B=\angle AA'D=90^\circ$. Hence $B,A',D$ are collinear.

Therefore, $A'$ is the foot altitude from $A$ to $BD$. Similarly, we get the properties of other points.

Now, $\angle BB'C=\angle BC'C$. Therefore, $B,B',C',C$ are concyclic.

So, $\angle PB'C'=\angle CB'C'=\angle CBC'=\angle CBP=\angle PBC$.

Similarly, $\angle PC'B'=\angle PCB$. 

Therefore, $\triangle PBC\sim\triangle PB'C'$. Similarly, $\triangle PAD\sim\triangle PA'D'$.

Since, $AA'\perp BD, CC'\perp BD$, hence $AA'\parallel CC"$. So $\frac{PC'}{PA'}=\frac{PC}{PA}$. 

Therefore, $ABCD\sim A'B'C'D'$.