$|R^2|$ alone having the distribution as above I have no problem,
however, I don't understand why the entire $Power_{rx}$ also have the
same distribution. At least, I see they multiple a bunch of constant
to R, shouldn't that will having some effect on the distribution as
well?
Yes, it does have an effect on the distribution -- the expectations are different, although the type of distribution is the same (i.e. Exponential).
You seem okay with the fact that if $R$ has a Rayleigh distribution, then $X := R^2$ has an Exponential distribution; i.e., the distribution of $X$ has the density function
$$f_{X}(x)=(1/(EX)) e^{-x/(EX)}(x \ge 0)
$$
and the cumulative distribution function is (by integration)
$$pr(X \le x) = (1 - e^{-x/(EX)})(x \ge 0)
$$
where $EX$ denotes the expectation of $X$.
But then $Y := c\cdot X$, where $c > 0$ is any constant (e.g., $c=Power_{tx}/(1+d^2)$), also has an Exponential distribution:
$$pr(Y \le y) = pr(c\cdot X \le y) = pr(X \le y/c) = (1 - e^{-(y/c)/(EX)})(y/c \ge 0) = (1 - e^{-y/(c\cdot EX)})(y \ge 0).
$$
That is, $c\cdot X$ also has an Exponential distribution, but with expectation $c \cdot EX$.
In more of your original symbols, the densities are
$f_{R^2}(x)=(1/X_{mean}) exp(-x/X_{mean})$, where $X_{mean}$ is $E(R^2)$, and
$f_{Power_{rx}}(p)=(1/P_{mean}) exp(-p/P_{mean})$, where $P_{mean}$ is $E(Power_{rx}) = E(R^2)\cdot Power_{tx}/(1+d^2)$.
