There are two answers to this question:1/3,5/3,

Take P on the left side of the arc as an example:

It is proved that △BHK is similar to △BGM and △BHK is similar to △HAO, and then BK= 1/3 is obtained by using some properties of the ratio.

The specific proof is as follows:

Solution: ∫ square ABCD, side length is 2, O is the midpoint of AD.

∴AO= 1,∠ABC=90

And ∵O is the center of the circle, OE is the radius, and the straight line MPG is the tangent of the circle O.

∴oh⊥mg ∴op⊥mpg

∴∠MPH=90 ∴∠BHP+∠BMG=90

In the right triangle BMG, ∠ bgm+∠ BMG = 90.

∴∠BHP=∠BGM

And because: ∠ HBK = ∠ GBM = 90 = ∠ A.

∴△BHK is similar to△ △BGM, and△ bhk is similar to△ Hao.

∴BG/BH=BM/BK，BK/AO=BH/AH

∴BG/BM=BH/BK=3,BH/BK=AH/AO=3

∴BH= 1,BK= 1/3

In the second case, when the point P is on the right side of the arc, it is similar to a triangle, and the corresponding sides are proportional. Then we can get it by using some properties of the ratio:

BK=5/3