Standardization typically means rescaling a random variable $X$ as
$$Z=\dfrac {(X-\mu_X)} {\sigma_X}$$
This is to give the standardized distribution $Z$ a mean of 0, and a standard deviation of 1 ($\mu_Z=0, \sigma_Z=1$). This is done to give us a common "yardstick" to compare/describe different distributions. See e.g. here.
There may be other definitions (e.g. rescaling to $[0,1]$, or even to $[0,100]$ -percentages), but I would argue that these definitions are inadvisable, as using the term standardized is confusing, and instead the term rescaled should be used.
Note that one can standardize any distribution (as long as it has a finite mean and standard deviation). One can of course standardize the normal distribution, but also a uniform one, an exponentional, a log-normal, etc. But one can not rescale all distributions, say, to $[0,1]$, certainly not a normal one, the support of which is $[-\infty,+\infty]$, or an exponential, etc. So this is another reason why I would call different definitions of "standardized" inadvisable.
Now, the reason classmates picked A) is the use of $Z$ as the name of the variable, because the standard normal distribution is almost always called $Z$. But not all standardized distributions called $Z$ need to be standard normal. Below is a standardized exponential distribution.

So, the correct answer is B). And the "giveaway" is the word always; it implies that, regardless of the original distribution of $X$, we will have
$${\sigma_Z}^2=E[(Z-\mu_Z)^2]=E[(Z-0)^2]=E[Z^2]=1$$
because standardization implies ${\sigma_Z}^2=\sigma_z=1$.