• A common method to prove quadrilaterals are similar is by dividing them into triangles and showing that corresponding triangles are similar.
• If you know about inversion, it is obvious.

$\angle AA'B=\angle AA'D=90^\circ$. 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, $ABCD\sim A'B'C'D'$.