Preliminary, let us do a slight change in notation: I will indicate with $k \in \mathbb{Z}$ the discrete time, with $K$ the $k$ present in the summation and the period (if it exists) with $N \in \mathbb{N}_0$.
Firstly, let us refresh the following concepts. Let $\theta \in \mathbb{R}_+$, $k \in \mathbb{Z}$, $i = \sqrt{-1}$, and consider the following (complex valued) function: $f(\theta) = e^{i \theta k}$. It is easy to check that this function is periodic with period $2\pi$, in fact
$
\begin{align}
f(\theta + 2\pi) = e^{i (\theta +2\pi) k} = e^{i \theta k} e^{i 2\pi k} = e^{i\theta k} = f(\theta)
\end{align}
$
since $e^{i 2 \pi k}=1$. Now, consider instead (with the same meaning of the symbols) the following (complex valued) function
$
\begin{align}
f(k) = e^{i \theta k}
\end{align}
$
Is it periodic? For discrete time functions, the period must be an integer, since the argument of the function must be an integer (the domain of $f(k)$ is $\mathbb{Z}$). A discrete time function can't be, e.g., periodic with period $\pi$. Thus, we must find, if it exists, an integer $N \in \mathbb{N}_0$ such as that $f(k+N) = f(k)$. Repeating the same reasoning as above,
$
\begin{align}
f(k+N) = e^{i\theta(k+N)} = e^{i \theta k} e^{i \theta N}
\end{align}
$
Now, $e^{i \theta N}$ is equal to $1$ if and only if $\ \theta N = 2 \pi n$, with $n \in \mathbb{N}$, which means:
$
\begin{align}
(*) \quad \theta = 2 \pi \frac{n}{N}, \quad n \in \mathbb{N}, N \in \mathbb{N}_0
\end{align}
$
The key concept is the following
The function $e^{i \theta k}$ is periodic (with respect to $k \in \mathbb{Z}$) if and
only if the pulsation $\theta \in \mathbb{R}_+$ satisfies $(*)$, i.e., the pulsation is a rational multiple of $\pi$.
Now, back on your question. Your example is unfortunately ill-worded, since the period of discrete time functions/processes can't be $2 \pi$ ("the last part is periodic with period $2 \pi$" is, strictly speaking, wrong): as explained above, only integers are allowed. In fact, since $k$ is used as an (integer) index in the formalism $X_k$, what means, e.g., $X_{\pi}$ ? There exists only things like $X_1, X_2, X_3, \dots$, because your process is discrete time (this is the reason why I asked and, as it's now clear, it is a crucial fact).
Luckily enough, this means that it is very easy to prove that, in general, $X(k)$ (i.e., the process viewed as a function of discrete time) is not periodic, since the phasors are in general not periodic (with respect to $k$). However, I doubt your professor was referring to this: he probably was referring to something along the lines of $f(\theta)$ is $2 \pi$-periodic, i.e., viewing the process as a function not of time, but of the pulsations $\lambda_j$.
At this point, it is useless to describe in detail the fact about the spectral distribution. Thus, I will just give you a sketch. Using the Wiener-Khinchin theorem, we know that the spectral density of a random process is equal to the Fourier transform of the ACF (AutoCorrelation Function) of the process. But the ACF of a periodic function is periodic (and with the same period), thus the information on periodicity is "carried on" the spectral density.
This is the same as what happens deterministically. Let $f(t)$, $t \in \mathbb{R}$, be a function of (continuous) time; the Fourier transform of $f(t)$, which we can denote with $F(\omega)$, where $\omega \in \mathbb{R}$ is the pulsation, contains all the information present in $f(t)$. In short, this is simply the usual time domain - frequency domain duality.
Example: This is a trivial example to show that the process is in general not periodic. Starting with
$
\begin{align}
X_k = \sum_{j=-K}^{K} A_j e^{i \lambda_j k}
\end{align}
$
Now, choose $\lambda_j = \lambda \ \forall j$ (single frequency process), from which
$
\begin{align}
X_k = \sum_{j=-K}^{K} A_j e^{i \lambda k} = e^{i \lambda k} \underbrace{\sum_{j=-K}^{K} A_j}_{=A} = A e^{i \lambda k}
\end{align}
$
The periodicity of $X_k$ depends on the periodicity of the phasor $e^{i \lambda k}$, which depends on $\lambda$, remember the condition $(*)$. Thus, we can easily conclude that $X_k$ is, in general, not periodic.
