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You really have to make a meal out of these redox equations.

Permanganate ion (STRONGLY COLOURED PURPLE) could be reduced to almost colourless $Mn^(2+)$....

$stackrel(+VII)[MnO_4]^(-) +8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_2O$ $(i)$

But here it is reduced to $MnO_2$.

$stackrel(+VII)[MnO_4]^(-) +4H^(+) + 3e^(-) rarr stackrel(+IV)"MnO"_2 + 2H_2O$ $(ii)$

Ammonia is OXIDIZED to nitrate anion.......(there is no associated colour change; ammonia pen and inks very badly, whereas ammonium ion/salt is odourless, but I do not suggest you use this method to differentiate them!)

$stackrel(-III)"NH"_3 +3H_2O rarr (stackrel(+V)"NO"_3)^(-) +9H^(+)+ 8e^(-)$ $(iii)$

Now if I have done my sums right; $(i)$, $(ii)$, and $(iii)$ ARE STOICHIOMETRICALLY BALANCED WITH RESPECT TO MASS AND CHARGE. If they are not, then they cannot be accepted as a representation of chemical reality. I think they are balanced, and so the final redox equation eliminates the electrons, and we take the sum...........$8xx(i)+5xx(iii)$:

$8MnO_4^(-) +5NH_3 + cancel(15H_2O) + cancel(64)19H^(+) rarr 8Mn^(2+) + 5NO_3^(-) +cancel(45H^(+))+ cancel(32)17H_2O$

And we cancel out common reagents to give (finally!):

$8MnO_4^(-) +5NH_3 +19H^(+) rarr 8Mn^(2+) + 5NO_3^(-) + 17H_2O$

Even despite the whack coefficients this is stoichiometrically balanced AS IS ABSOLUTELY REQUIRED...........

But you specified the reduction of permanganate to $MnO_2$, so this is ...........$3xx(i)+5xx(iii)$:

$8MnO_4^(-) +3NH_3 +5H^(+)rarr 8MnO_2(s) + 3NO_3^(-) +7H_2O$

But I note (well eventually I did!) that you specified BASIC conditions; all I have to do is add $5xxHO^-$ to each side of the equation............and eliminate the waters..........

$8MnO_4^(-) +3NH_3 rarr 8MnO_2(s) + 3NO_3^(-) +5HO^(-) + 2H_2O$

Manganese metal has a particularly rich redox chemistry. For another example of this redox manifold, see See

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