$4$ bits are for the opcode so number of $2$ address instructions will be $2^4 =16-$ so $14$ double instructions are possible.
But out of $16$ only $14$ are used so $2$ are still left which can be used for $1$ address instruction. For $1$ address instruction we can use not only the $2$ left over but also the $6$ bits of operand $1$ (to make it one address) $-$ so $6$ bits that is $64.$ So, total $2\times 64$ single address instructions can be supported $-$ So, $127$ single instructions are possible
But out of $128,127$ are used so $1$ left which can be used for zero-address instruction. To make number of zero address we can use the operand $2$ address (we already included operand 1 address) $- 6$ bits. So, total number possible is $64.$ So,
total $1\times 64 = 64$ zero address instructions are possible.
So, all encoding are possible.