Could someone take a look at this please?

Questions like this, no given lengths, just ratios, just about always, (probably always ?), involve similar triangles.

So, look for triangles that are similar.

Above of the line PB, the angle APB  (APD) is common to two different triangles.

Angle PAD is equal to angle ABD. This comes from a circle/tangent/chord theorem. If you are not familiar with it, make a point of at looking at its proof, and then remember it.

That's enough, you now have two similar triangles.

Write down the ratios of the sides,

Repeat for triangles below PB. That's easy, all you have to do is replace A from the set of ratios you already have with C.

Anything else ? 

Yes, PA = PC. That allows you to bring the two sets of ratios together.

Now follow your nose.

(BTW, I think that this same question was asked some months ago. Probably there is a complete solution given back then.)

Here is the original question that guest is talking about.

https://web2.0calc.com/questions/geometry-proof_2

 Thanks CPhill and Tiggsy      

I had it on my watch list because it was so tricky!