Let's first write down a density for a scale-shape parameterization for a Gamma and then shift it. Taking the density from Wikipedia (which has it correct), but making the variable $z$ rather than $x$:
$$f(z;\alpha ,\theta )=\frac{z^{\alpha -1}e^{-z/\theta }}{\theta ^{\alpha }\Gamma (\alpha )}\quad {\text{ for }}z>0,\quad \alpha ,\theta >0$$
Translating to the parameterization and notation your book is using (aside from the variable):
$$f(z;p ,a )=\frac{z^{p -1}e^{-z/a }}{a ^{p }\Gamma (p )}\, \mathbb{1}_{[0,\infty]},\quad p,a >0$$
You can immediately see that the book has the ordinary gamma wrong by simply putting $A=0$, and discovering they're missing a factor of $a$ on the denominator. This arises because of the Jacobian when the scale parameter is introduced; if you omit it the thing doesn't integrate to 1. You should check my assertion for yourself by integrating it (take it back to the form of an ordinary gamma integral, which you know the value of).
Now let's consider what happens when shift it up by $\delta$ (their $A$). Let $X = Z+\delta$, so $Z=X-\delta$ and $dx = dz$. The density therefore becomes:
$$f(x;p ,a ,\delta)=\frac{(x-\delta)^{p -1}e^{-(x-\delta)/a }}{a ^{p }\Gamma (p )}\, \mathbb{1}_{[\delta,\infty]},\quad p,a >0,\: \delta \in \mathbb{R}$$
Which now differs from the book in several respects. The book is wrong. (A little experience with location-scale families will tell you so at a glance.)
If you make sure to start with a correctly specified density (i.e. by starting with the phrase "shifted gamma" and working out what the density of that must be), you will get more out of the exercise. Don't rely on the book to be correct.