Inverse Fourier Transform Dirac impulse with scaled argument
This is just a matter of variable substitution in the integral. The inverse Fourier transform of $\delta(\Omega T)$ is

$$\mathcal{F}^{-1}\big\{\delta(\Omega T)\big\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\Omega T)\,e^{j\Omega t}\,d\Omega\tag{1}$$

Substituting $\alpha=\Omega T$ gives $d\alpha=d\Omega\cdot T$, and $(1)$ becomes

$$\mathcal{F}^{-1}\big\{\delta(\Omega T)\big\}=\frac{1}{2\pi T}\int_{-\infty}^{\infty}\delta(\alpha)\,e^{j\alpha t/T}\,d\alpha=\frac{1}{2\pi T}\tag{2}$$

The last equation is a consequence of the sifting property of the Dirac delta impulse.