BPSK vs $\pi/2$ BPSK, what the benefits?
The signal always changes phase at bit transition times. This makes it easier to achieve bit synchronism at the receiver. In plain BPSK, long runs of 0 or 1, with no phase transitions occurring at the boundary
between two identical bits, the phase-lock loop that is generating the
signal telling the receiver where the bit transitions are, can lose lock; or the initial acquisition is more difficult. Also, in satellite channels with nonlinear amplifiers such as traveling-wave-tube (TWT)
amplifiers, smaller phase transitions are better. In particular, if the RF carrier phase makes
a transition of $\pi$ radians (meaning the carrier envelope makes a
transition through the origin), the TWT amplifier output contains
all the sidebands that were carefully suppressed in the uplink. So,
filtering in the satellite is required to suppress these sidebands.
Hence, $\pi/2$-BPSK and $\pi/4$-QPSK are preferred over their plain cousins. Note 4, not 2, with QPSK: $\pi/2$-QPSK does not avoid the
transitions through $0$, and adds complexity without necessarily
adding benefits.

Note: in plain BPSK, the signals are $\pm\cos(\omega_0 t)
= \{\cos(\omega_0 t), \cos(\omega_0t + \pi)\}$, and so,
at bit transitions, the phase either stays the same, or changes by
$\pi$. in $\pi/2$-BPSK, the signal set is $\pm\cos(\omega_0 t)$
for one bit and $\pm\sin(\omega_0 t)$ for the next. Thus,
regardless of whether we have two successive identical bits
or two different successive bit, the signal phase changes by $\pm \pi/2$
since the signal changes from a cosine to a sine.  Figure out for
yourself why $\pi/2$-QPSK has transitions through $0$ while
$\pi/4$-QPSK does not; then think of offset QPSK a little.