Yeah, you could be the greatest

Because PC is tsngent, and angle BCD is 90 degrees, angle PCD is 135 degrees, and PCB is 45. Also, we can find the squares side lengths to be $\sqrt{2}$ . Because PC is tangent to the square, and PCD is 135 degrees, we know triangle BCP is a 45,45,90 with side lengths root 2. Using the Pythagoras theorem: $2+2=4$ . The square of which is 2. So PC has a length of 2. We know two lengths and the angle, so we can use the law of cosines: $a^{2} = \sqrt {2}^{2}+2^{2}-2 (2)(\sqrt {2})*cos (135)$ . Solving for a, we get $\sqrt {10}$ .