Why is $(\sqrt{P})^2=P$ where $P$ is a positive operator on a Hilbert space?

Question

The following is a proposition regarding positive operators on a Hilbert space in Douglas's Banach Algebra Techniques in Operator Theory:

Corollary 4.32 is as the following:

I understand that the (continuous) functional calculus is defined as the following:

Here is my question:

How is it true that $(\sqrt{P})^2=P$ by the definition of the functional calculus?