Batman tap-dances!

We need the minimum vaue for $B+M+T$ . We can assign $1,2,3$ in any way possible. We'll put those possibilities on hold and look at the other part. Now, we need to minimize the value of $A$ using digits $0,4,5...9$ such that $3A$ is minimum. The best option would be to choose $A=0$ . Now, We need to minimise $T+N+ P$ using values from $4,5...9$ . But we already have 3 possible values for $T$ , of which the minimum is $1$ .

Summarizing,

$A = 0$

$B = 2 \space or \space 3$

$M = 2 \space or \space 3$

$N = 4 \space or \space 5$

$P = 4 \space or \space 5$

$T = 1$

The possibilities of values will not affect the overall sum, since each occur under the same place value.

Put A=0(since first letter is non zero,making the second letter zero is the best way.T occurs twice.so give T=1.M=2,B =3 or vice versa.P=4,M=5 or vice versa. BAT-301 MAN-204 TAP-105 Sum-610