Finding the correct $x[n]$ from a given DFT $x[k]$
Let's solve the exercise. It's asking for your awareness of periodicity in DFT and some trigonometry.

First, beacuse of the question content, let's state the following :
$$\delta[n-d] \overset{DTFT} \longleftrightarrow e^{-j \omega d } \overset{DFT} \longleftrightarrow e^{-j \frac{2 \pi}{N} d k } $$ 

Then given our DFT $X[k]$ work it backwards into $x[n]$ as a sum of delayed impulses.

$$ 
\begin{align} 
X[k] &amp;= 2 \cos(\frac{\pi k}{2}) + 2 \sin(\frac{14 \pi k}{8}) + \sin(\pi k) \\ 
X[k] &amp;= 2 \cos(\frac{\pi k}{2}) + 2 \sin( \left( 2\pi - \frac{2\pi}{8} \right) k) + 0 \\ 
X[k] &amp;= 2 \cos(\frac{\pi k}{2}) + 2 \sin(- \frac{2\pi}{8} k) \\ 
X[k] &amp;= 2 \cos(\frac{\pi k}{2})  - 2\sin(\frac{\pi}{4} k) \\ 
\end{align}
$$

We have used basic trigonometry to reach the above line. Now lets use Euler's formula to expand the cosines and sines into complex exponentials so that we can use the corresponence of the first line to deduce the time domain impulses.

$$ 
\begin{align} 
X[k] &amp;= 2 \cos(\frac{\pi k}{2})  - 2\sin(\frac{\pi}{4} k) \\ 
X[k] &amp;= 2 \left( \frac{ e^{j \frac{\pi}{2} k} + e^{-j \frac{\pi}{2} k} }{2} \right)  -2 \left( \frac{ e^{j \frac{\pi}{4} k} - e^{-j \frac{\pi}{4} k} }{2j} \right) \\
X[k] &amp;= \left( e^{j \frac{\pi}{2} k} + e^{-j \frac{\pi}{2} k}  \right)  +j \left( e^{j \frac{\pi}{4} k} - e^{-j \frac{\pi}{4} k} \right) \\  
\end{align}
$$

Now the correspondence between $\delta[n-d]$ and its $N=8$ point DFT is:
$$\delta[n-d] \overset{DTFT} \longleftrightarrow e^{-j \omega d } \overset{DFT} \longleftrightarrow e^{-j \frac{2 \pi}{8} d k } = e^{-j \frac{\pi}{4} d k } $$ 

And by using ths relationship we can deduce from the last line that the impulses are:

$$ 
\begin{align} 
X[k] &amp;= e^{j \frac{\pi}{2} k} + e^{-j \frac{\pi}{2} k}  +j e^{j \frac{\pi}{4} k} - e^{-j \frac{\pi}{4} k} \\  
x[n] &amp;= \delta[n+2] + \delta[n-2] + j\delta[n+1] -j \delta[n-1] 
\end{align}
$$

One last step remains to complete the answer; representation of $x[n]$ in the range $0 \leq n \leq 7$ as it is the valid range for the formal relationship between $x[n]$ and its $N$-point DFT $X[k]$. Then we should map those impulses outisde of this range into it. As you can see $\delta[n+1]$ and $\delta[n+2]$ are outside of this range and should be mapped into $0 \leq n \leq 7$ by shifting them right by $N=8$. This shift is the consequence of the inherent periodicity in both $x[n]$ and $X[k]$ as $x[n-N] = x[n]$ etc. So we decude that $x[n]$ in the range $0 \leq n \leq 7$ is:

$$x[n] = \delta[n+2-8] + \delta[n-2] + j\delta[n+1-8] -j \delta[n-1]$$
$$x[n] = \delta[n-6] + \delta[n-2] + j\delta[n-7] -j \delta[n-1]$$

which is shown in the option b.