Step 1: Understanding the given condition.
The condition
\[
P\left(\frac{a}{2\pi} X \in \mathbb{Z}\right) = 1
\]
means that \( X \) takes values only at discrete points of the form \( X = \frac{2\pi n}{a} \), where \( n \) is an integer. Hence, \( X \) is a discrete random variable supported on multiples of \( \frac{2\pi}{a} \).
Step 2: Characteristic function definition.
The characteristic function of \( X \) is defined as:
\[
\phi_X(t) = E[e^{itX}] = \sum_{n=-\infty}^{\infty} p_n e^{it \frac{2\pi n}{a}},
\]
where \( p_n = P\left(X = \frac{2\pi n}{a}\right) \).
Step 3: Compute \( \phi_X(a) \).
Substituting \( t = a \), we get:
\[
\phi_X(a) = \sum_{n=-\infty}^{\infty} p_n e^{i a \frac{2\pi n}{a}} = \sum_{n=-\infty}^{\infty} p_n e^{i 2\pi n} = \sum_{n=-\infty}^{\infty} p_n (1) = 1.
\]
Thus, \( \phi_X(a) = 1 \), which makes statement (A) true.
Step 4: Checking periodicity.
Now, consider:
\[
\phi_X(t + a) = \sum_{n=-\infty}^{\infty} p_n e^{i (t + a) \frac{2\pi n}{a}} = \sum_{n=-\infty}^{\infty} p_n e^{i t \frac{2\pi n}{a}} e^{i 2\pi n} = \sum_{n=-\infty}^{\infty} p_n e^{i t \frac{2\pi n}{a}} = \phi_X(t).
\]
Thus, \( \phi_X(t) \) is periodic with period \( a \), proving statement (B) true.
Step 5: Analysis of other options.
(C) \( |\phi_X(t)|<1 \) for all \( t \neq a \) is not necessarily true because for a discrete distribution, \( \phi_X(t) \) may attain 1 at other multiples of \( a \).
(D) The integral expression does not hold in this general setting—it is unrelated to the periodicity condition and characteristic function form.
Step 6: Conclusion.
Therefore, both statements (A) and (B) are correct.