Solution:

We know $19!$ is a multiple of $5^3$ and $2^3,$ so it is a multiple of $1000.$ Therefore, $H=0$

We know it is a multiple of $9,$ so its digit sum must be a multiple of $9.$ As such, $$1+2+1+6+T+5+1+$$$$0+0+4+0+M+8$$$$+3+2+0+0+0$$$$=33+M+T$$ is a multiple of $9.$ With this in mind, we know that $$M+T \equiv 3 \mod 9,$$ leaving just $3$ and $12.$

We also know it is a multiple of $11,$ which means that when alternating between adding and subtracting digits, we get $$1-2+1-6+T-5+1-$$$$0+0-4+0-M+8-$$$$3+2-0+0-0$$$$= T-M-7$$ is a multiple of $11,$ so $$T-M \equiv 7 \mod 11.$$

The only way to satisfy both is $$T=4,M=8,H=0.$$ Their sum is $12.$

Thus, the answer is C .