The membership problem $A_{LBA}$ for a deterministic $LBA$ is decidable because the number of configurations that a $LBA$ can assume is finite. Since this number is also finite for a non-deterministic $LBA$, how can this information be used to show that the membership problem $A_{LBA}$ is decidable for a non-deterministic $LBA$?

Thanks in advance, Marcus.

The membership problem $A_{LBA}$ for a deterministic $LBA$ is decidable because the number of configurations that a $LBA$ can assume is finite. Since this number is also finite for a non-deterministic $LBA$, how can this information be used to show that the membership problem $A_{LBA}$ is decidable for a non-deterministic $LBA$?

Thanks in advance, Marcus.

P.S. Suppose the $LBA$ has $|Q|$ states, the size of the input is $|w|$ and the size of the alphabet is $|\Gamma|$. Then the number of possible configurations for this $LBA$ is $|Q|*(|w|+2)*|\Gamma|^{|w|}$. So it is finite.