* What is the length of DP?
* What is the angle APB?

For the first part, draw lines parallel to the sides of the square through P. Let the segment parallel to AD intersect AB at E and the segment parallel to AB intersect AD at F. There's an intersection at the other sides too but it's not necessary to discuss it since it breaks the other sides into the same lengths as these sides. Let AE=w, BE=x, AF=y, DF=z.

Applying the Pythagorean theorem 4 times,

AP^2=w^2+y^2 = 1
BP^2=x^2+y^2 = 4
CP^2=x^2+z^2 = 9
DP^2=w^2+z^2

You can get w^2+z^2 by adding the first and the third equation and then subtracting the 2nd to arrive at w^2+z^2=1+9-4=6. Therefore, DP^2=6 --> DP=sqrt(6).

As for the 2nd part, I could not get an analytical solution so I used Excel Goal Seek. It would be nice if someone can come up with an analytical solution.

Let APB be J, BPC be K, CPD be L, and DPA be M. Also, let the side of the square be length s. Apply the law of cosines 4 times yields

s^2=5-4cosA
s^2=13-12cosB
s^2=15-6sqrt(6)cosC
s^2=7-2sqrt(6)cosD
Also, A+B+C+D = 2pi

We have 5 equations and 5 unknowns. I couldn't come up with an analytical solution. So I defined cosA, cosB, cosC in terms of s^2. Then I found A, B, and C and used the fifth equation to find D. I used Excel Goal Seek to find s^2 such that the fourth equation applies. The results I got are:

s=2.797921, A=2.356172, B=1.125229, C=1.061036, D=1.740749. (Radians). The required angle is A (135 degrees). Since it has such a nice answer, I assume it can be done analytically but I have no idea how it's done.