CTFT to DTFT why can't you always just substitute $\Omega = \omega/T_S$ This is something I've always wondered about in DSP class, but just accept as a fact because I never really fully understand why this is the case:

Given CTFT:

$$X_s(j\Omega) = 6000 \pi \sum \limits_{k=-\infty}^{\infty} \Bigg[ \delta(\Omega - 1200\pi k - 4000 \pi)  
 ~+~  \delta(\Omega - 1200\pi k + 4000 \pi)\Bigg]$$

Then converting to DTFT using formula:

$$X(e^{j\omega})=\frac{1}{T_s} X_s\bigg(\frac{\omega}{T_s}\bigg)$$

But, Why is answer this:

$$X(e^{j\omega}) = 6000 \pi \sum \limits_{k=-\infty}^{\infty} \Bigg[ \delta(\omega + T_s(-1200\pi k - 4000 \pi))  
 ~+~  \delta(\omega +T_s(- 1200\pi k + 4000 \pi))\Bigg]$$

instead of this:

$$X(e^{j\omega}) = 6000 \pi \sum \limits_{k=-\infty}^{\infty} \Bigg[ \delta(\frac{\omega}{T_s} - 1200\pi k - 4000 \pi)  
 ~+~  \delta(\frac{\omega}{T_s} - 1200\pi k + 4000 \pi)\Bigg]$$



seems like this formula: $$X(e^{j\omega}) = \frac{1}{T_s} X_s\bigg( \frac{\omega}{T_s}\bigg)$$ should really be:  $$X(e^{j\omega}) = \frac{1}{T_s} X_s\bigg( \omega = \Omega T_s\bigg)$$
It is due to the property of unit impulse function

\begin{equation}
\delta(\alpha t)  = \frac{ 1}{|\alpha|} \delta(t) 
\end{equation}

(Usually $\omega $ is used for angular frequency and $\Omega $ is used for angle. Interpret my answer accordingly. )

suppose 
\begin{equation}
X_a(\omega) = \delta(\omega -x_0 )  
\end{equation}
and if we substitute '$\omega$' with $\frac{\Omega}{T_s}$ then 
\begin{equation}
X_a(\frac{\Omega}{T_s}) = \delta(\frac{\Omega}{T_s}  -x_0 )
   = \delta(\frac{\Omega - x_0 T_s}{T_s} )
   =  \delta(\frac{ 1}{T_s} (\Omega - x_0 T_s) )
   =  {T_s} *\delta(\Omega - x_0 T_s) 
\end{equation}
where '$\alpha$'  is $\frac{ 1}{T_s}$ 
and 
$$
\frac{ 1}{|\alpha|} = T_s 
$$

and when you do DTFT calculation on 

\begin{equation}
X(\Omega)  = \frac{ 1}{T_s} X_a(\frac{\Omega}{T_s})
   =\frac{ 1}{T_s} * ( T_s * \delta(\Omega - x_0 T_s) )
   = \delta(\Omega - x_0 T_s) 
\end{equation}

in the final answer $T_s$ and $\frac{ 1}{T_s}$ cancels 

and you will get the expression based on the independent variable '$\Omega - x_0 T_s$'

So there is nothing wrong in your actual answer.