Let $f$ be a real number, where $f=\dfrac{a-ib}{di}$, then
$f=\dfrac{a-ib}{di}*\dfrac{-di}{-di}=\dfrac{-adi-db}{d^2}$
As $f$ is a real number, $f=\dfrac{-db}{d^2}=\dfrac{-b}{d}$
On the other hand, we can say that $\dfrac{1}{f}=\dfrac{di}{a-ib}$ hence,
$\dfrac{1}{f}=\dfrac{di}{a-ib}*\dfrac{a+ib}{a+ib}=\dfrac{adi-bd}{a^2+b^2}$
Hence as $f$ is real, then $f=\dfrac{a^2+b^2}{-bd}$
I am wondering which one is correct.
Both of your expressions are correct, and are equivalent to each other. If $f$ is a real number, then you have $a = 0$ in both cases, so the $i$ factor in the denominator cancels with it in the numerator of $f = \frac{a - ib}{di}$. You can also see this when, for both case, you used that if $f$ is a real number, then $adi = 0$, but since $d \neq 0$ as it's in the denominator, you must have $a = 0$ instead.
Thus, in your second case, you get
$$f = \frac{a^2 + b^2}{-bd} = \frac{b^2}{-bd} = \frac{-b}{d} \tag{1}\label{eq1A}$$
This, of course, matches what you got earlier.