It's not necessary for a discrete-time signal to have the same period as the continuous-time signal that it represents. As long as the continuous-time signal is band-limited and as long as the sampling frequency is greater than twice the bandwidth of the continuous-time signal, the discrete-time signal can perfectly represent the continuous-time signal.
In your example, the continuous-time signal is
and the discrete-time signal is
$x_d[n]$ perfectly represents $x(t)$ if $f_0T<\frac12$ is satisfied. In general, $x_d[n]$ is not even a periodic sequence. It is only periodic if $f_0T$ is rational.
With the choice $f_0T=3/16$ we make sure that $x_d[n]$ perfectly represents $x(t)$, and we also make sure that $x_d[n]$ is periodic with period $16$. Note that there can't be $3$ periods of the discrete-time signal in the DFT frame, simply because $16/3$ is not an integer.
Since $x_d[n]$ has period $16$, taking a length $16$ DFT results in a single frequency component. With $f_0T=3/16$, that component occurs at frequency index $k=\pm 3$, which exactly represents a continuous-time sinusoid at frequency $3/16\cdot f_s$ (with $f_s=1/T)$.