Hello, neno89!
Recall this theorem:
If two secants are drawn to a circle from an external point,
the included angle is one-half the differnce of the intercepted arcs.
In your diagram: \(\displaystyle \,\angle P\;=\;\frac{1}{2}\left(\overline{WX}\,-\,\overline{ZY}\right)\)
We are told that: \(\displaystyle \,\overline{WXY}\:=\:200^o\;\;\Rightarrow\;\;\overline{WX}\,+\,\overline{XY}\:=\:200^o\;\;\) [1]
Then: \(\displaystyle \,\overline{WZY}\:=\:160^o\;\;\Rightarrow\;\;\overline{WZ}\,+\,\overline{ZY}\:=\:160^o\;\;\) [2]
Subtract [2] from [1]: \(\displaystyle \,(\overline{WX}\,+\,\overline{XY})\,-\,(\overline{WZ}\,+\,\overline{ZY})\:=\:40^o\)
We have: \(\displaystyle \,\overline{WX}\,+\,\overline{XY}\,-\,\overline{WZ}\,-\,\overline{ZY}\:=\:40^o\)
\(\displaystyle \;\;\)But \(\displaystyle WZ\,=\,XY\), hence: \(\displaystyle \,\overline{WZ}\,=\,\overline{XY}.\;\;\) (Equal chords subtend equal arcs)
Our equation becomes: \(\displaystyle \overline{WX}\,-\,\overline{ZY}\:=\:40^o\) . . . the difference of the intercepted arcs!
Therefore: \(\displaystyle \,\angle P\:=\:\frac{1}{2}(40^o)\:=\:20^o\)
