$x[n]$ after sampling of $cos(16\pi t+\phi)$ at 12kHz

I'm not sure what the question really means, so this is just guesswork.

I think options 1 and 4 can be ruled out as $$w_0<\pi$$.

The CTFT of $$cos(16\pi t+\phi)$$ has two spikes at $$16\pi$$ and $$-16\pi$$. I guess that the question means that we sampled this spectrum at $$12Hz$$, (because 12kHz doesn't make sense to me, might be a typo in the question), i.e. at $$24\pi$$.

So we get one copy of $$-16\pi$$ at $$-16\pi+24\pi=8\pi$$. Similarly we get a copy of $$16\pi$$ at $$-8\pi$$.

Now we normalize to the $$w$$ by diving by $$12$$. Only the frequencies $$-8\pi$$ and $$8\pi$$ get mapped in the range $$-\pi$$ yo $$\pi$$ as required by the question. These frequencies get mapped at $$-\frac{2}{3}\pi$$ and $$\frac{2}{3}\pi$$. But the problem is that they also get scaled by $$12Hz$$ during sampling (and I'm ignoring that to get to one of the options). So the $$x[n]$$ corresponding to $$\frac{2}{3}\pi$$ and $$frac{2}{3}\pi$$ is $$cos(\frac{2}{3}\pi n+\phi)$$

Is this correct?

First of all, as you said the sampling rate is probably $$12$$ Hz, rather than $$12$$ kHz, and perhaps they want to demonstrate an aliasing example.
Given a bandlimited continuous-time periodic signal $$x(t) = \cos(16\pi t + \phi)$$ the samples taken at the rate $$F_s = 12$$ Hz will be denoted as $$x[n]$$ and will be obtained by via $$x[n] = x(t_n)$$ with $$t_n = n T_s = n/Fs$$ :
$$x[n] = \cos( 16 \pi \frac{1}{12} n + \phi) = \cos(\frac{4 \pi}{3} n + \phi)$$
$$x[n]$$ is a discrete-time periodic sequence with frequency $$w_0 = \frac{4\pi}{3}$$, however, in the discrete-time case we are interested in the range of frequencies that fall in $$[-\pi,\pi)$$, and therefore this signal's frequency will be cast as:
$$x[n] = \cos(\frac{4 \pi}{3} n + \phi) = \cos(\frac{2 \pi}{3} n + \phi)$$
which indicates that the original continuous-time frequency of $$8$$ Hz is lost, and a new aliased frequency of $$4$$ Hz is attained...