No, there's very little loss from the atmosphere at 432 MHz. The 390 dB figure is a misunderstanding of how the path loss is calculated. The poster "wl" took the one way path loss (195 dB) and doubled it. But the moon is a huge reflector and provides "gain".

Here's a C program with the correct equation.

  #include <stdio.h>
  #include <stdlib.h>
  #include <math.h>

  int main(int argc, char **argv)
  {
    double  d, f, lambda, loss;

    if (argc != 3) {
      fprintf(stderr, "usage: moon <km> <frequency(MHz)>\n");
      exit(-1);
    }

    d = atof(argv[1]) * 1000.0;
    f = atof(argv[2]);

    lambda = 299792458.0 / (f * 1000000.0);
    loss = (0.065 * (1.738e6 * 1.738e6) * (lambda * lambda)) / (631.65468167 * (d * d * d * d));
    printf("EME path loss = %f dB\n", 10 * log10(loss));
    return 0;
  }

1.738e6 is the radius of the moon in meters and 0.065 is the reflection efficiency.

Satellites make moonbounce impracticable.