Proof regarding the periodicity of a continuous-time sinusoid after sampling Question
A continuous-time sinusoid $x_a(t)$ with fundamental period $T_p = \frac{1}{F_0}$ is sampled at a rate $F_s = \frac 1 T$ to produce a discrete-time sinusoid $x(n) = x_a(nT)$.


Show that $x(n)$ is periodic if $\frac{T}{T_p} = \frac K N$ (that is, $\frac{T}{T_p}$ is a rational number)
If $x(n)$ is periodic, what is the fundamental period $T_p$ in seconds?


I feel there is some ambiguity in the question itself. I assume $N$ here refers to the period of the sampled discrete-time signal.

I know how to prove that a discrete-time signal is periodic only if its relative frequency is a rational number. But the concept of periodicity in both continuous-time as well as sampled signal confused me.

So here is what I have tried:

Consider a continuous time sinusoid, $$x_a(t) = \cos(2 \pi Ft) \tag{1}$$ For this signal to be continous with a fundamental period $T_p$, it must follow that $$F=\frac{k_1}{T_p} \tag{2}$$ for some $k_1$ which is an integer.

Now that this signal is sampled with samppling rate $F_s$, the resulting discrete-time signal will be:
$$x(n) = \cos(2\pi f n)\tag{3}$$ where, $$f = \frac{F}{F_s} \tag{4}$$ This sampled signal will be periodic with period $N$ iff
$$f = \frac{k_2}{N} \tag{5}$$ for some $k_2$ which is also an integer.

Using equations (2), (4) and (5), it follows
$$\frac{T}{T_p} = \frac{k_2}{k_1 N} \tag{6}$$ If I redefine $\frac{k_2}{k_1}$ to be $K$, then equation (6) becomes
$$\frac{T}{T_p} = \frac{K}{N} \tag{7}$$ which proves the first part.

Now that the fundamental period $T_p$ has been asked, solving for $T_p$ in equation (6) gives
$$T_p = \frac{k_1 T N}{k_2} \tag{8}$$

Since both discrete time period and continuous time periods are in the same place, this has made me confused whether my approach to the solution is correct or not. Could you please verify that the approach is correct? Or if anything is wrong, could you point that out to me? Thanks.
You might be making this problem harder than it is by thinking in terms of frequency rather than periods.  Your sinusoid is
$$x(t) = \cos\left(2\pi \frac{t}{T_p}\right)$$ and its samples are
$$x[n] = x(nT) = \cos\left(2\pi \frac{nT}{T_p}\right).$$
The samples form a discrete-time periodic sequence of period $N$ if 
\begin{align}
x[n] &amp;= x[n+N]~\forall\, n\\
&amp;{\Large \Downarrow}\\
\cos\left(2\pi \frac{nT}{T_p}\right) &amp;= \cos\left(2\pi \frac{(n+N)T}{T_p}\right)\\
&amp;{\Large \Downarrow}\\
\cos\left(2\pi \frac{nT}{T_p}\right)&amp;= \cos\left(2\pi \frac{nT}{T_p} + 2\pi \frac{NT}{T_p}\right)
\end{align}
where the last equality will hold only if $\displaystyle\frac{NT}{T_p}$ equals some integer $k$. This is equivalent to the OP's desired requirement that $$\frac{T}{T_p} = \frac kN$$ for some integers $k$ and $N$.

So, what is the fundamental period of $x[n]$? That is,
what is the smallest integer $N_0$ such that $\displaystyle\frac{N_0T}{T_p}$ is an integer?  Well, suppose that the ratio $\displaystyle\frac{T}{T_p}$ is the rational number $\frac{k_1}{k_2}$ where $k_1$ and $k_2$ are relatively prime integers. Then, $N_0 = k_2$ and $\displaystyle\frac{N_0T}{T_p} = k_1$.  Thus, considering $x[n]$ as a periodic discrete-time sequence of samples spaced $T$ seconds apart  -- think of it as the impulse train $\sum_n x(nT)\delta(t-nT)$ -- gives a (continuous-time) period of $k_2T$ seconds for the sequence/impulse train. Note that one period of the discrete-time sequence spans $k_1$ periods of the sinusoid of period $T_p$.