A number written in base ten is a multiple of $4$ if and only if the two-digit number formed by its last two digits is a multiple of $4$. If these two digits are $ab$ in that order, clearly $b$ must be even. It’s not hard to verify that if $b$ is a multiple of $4$ (i.e., $0,4$, or $8$), then the two-digit number $ab$ is a multiple of $4$ if and only if $a$ is even, while if $b\equiv2\pmod 4$ (i.e., $2$ or $6$), then $ab$ is a multiple of $4$ if and only if $a$ is odd.
Now suppose that you have your four-digit number $n$ whose permutations are all multiples of $4$. Clearly $n$ cannot contain an odd digit, so it also cannot contain a digit $2$ or $6$. We conclude that all four digits must be multiples of $4$, i.e., must be $0,4$, or $8$. Now just count the four-digit numbers that can be formed from these three digits, being careful to exclude leading zeroes.