If the radiation pressure a distance $d>R$ from the center of an isotropic black body star is found to be $P_{rad}=\large{\frac{4\sigma T^4}{3c}}\left[1-\left(1-\frac{R^2}{d^2}\right)^{\frac{3}{2}}\right]$
a) How do I show that $P_{rad}$ obeys an inverse square law for $d>\!\!>R$?
b) Why does the inverse square law scaling break down close to the stars surface?
This procedure is used very often in physics in order to simplify calculations and make easier the physical interpretation when we have limited measurement accuracy or we don't need a high degree of accuracy (or we need to explain some point about conflicting theories)
First, consider the series expansion of $(1-x^2)^\frac{3}{2}$
$$(1-x^2)^\frac{3}{2}=1-\dfrac{3}{2}x^2+\dfrac{3}{2}x^4+\dfrac{1}{16}x^6+...$$
Now, if $0<x\approx 0$ we can approximate the LHS by $1-\dfrac{3}{2}x^2$ (we drop the square, the cube, etc. of a value almost zero)
Now, define $x=\dfrac{R}{d}$ and $W=\dfrac{4\sigma T^4}{3c}$
$$P_{rad}=W\left[1-\left(1-x^2\right)^{\frac{3}{2}}\right]$$
But $R<\!\!<d$, then $0<x\approx 0$ too. Then
$$P_{rad}\approx W\left[1-\left(1-\dfrac{3}{2}x^2\right)\right]=\dfrac{3WR^2}{2}\dfrac{1}{d^2}$$
Which is an inverse square law.
Finally, if we are considering points near the star, we have $R\approx d$, we have to include many terms from the series expansion and, as they depend on the inverse of the fourth, sixth, ... power of $d$, this obviously makes the formula very different from the simple one the inverse square law has.