= Solution
Take positive <velocity> towards $B$, and interpret $100\,\mathrm{MHz}$ as the emitted <frequency> in the probe's rest frame. The <relativistic velocity-addition formula> gives the probe's station-frame <velocity>:
$$
u_P=\frac{0.5c+0.5c}{1+(0.5)(0.5)}=0.8c.
$$
For reception at the station, photons propagate backwards while the probe recedes. The longitudinal <relativistic Doppler effect> gives
$$
\boxed{f_{\rm station}=100\sqrt{\frac{1-0.8}{1+0.8}}\,\mathrm{MHz}
=\frac{100}{3}\,\mathrm{MHz}\simeq33.33\,\mathrm{MHz}.}
$$
Before the probe overtakes $B$, their separation decreases. The probe's <velocity> in $B$'s <inertial frame> is
$$
u_{P|B}=\frac{0.8c-0.4c}{1-(0.8)(0.4)}=\frac{10}{17}c.
$$
The approaching-source <relativistic Doppler effect> therefore gives
$$
\boxed{f_B=100\sqrt{\frac{1+10/17}{1-10/17}}\,\mathrm{MHz}
=100\sqrt{\frac{27}{7}}\,\mathrm{MHz}\simeq196.40\,\mathrm{MHz}.}
$$
An independent frame calculation gives the same result: forward photons have station-frame <frequency> $300\,\mathrm{MHz}$, and a receiver moving at $0.4c$ measures $300\sqrt{0.6/1.4}\,\mathrm{MHz}$. If reception occurs after the probe passes $B$, backwards photons instead give $f_B=100\sqrt{7/27}\,\mathrm{MHz}\simeq50.92\,\mathrm{MHz}$. The direction of propagation and the reception geometry are essential to choosing the <Doppler factor>.
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