Given the set P = 1, 3, 5, 6 and the operation a * b = (ab + 3) mod 8, complete the operation table.

The operation $*$ is defined as $a * b = (ab + 3) \pmod{8}$. This means we multiply $a$ and $b$, add 3, and then find the remainder when divided by 8. The set is $P = \{1, 3, 5, 6\}$.

We will calculate $a * b$ for each pair $(a, b)$ from the set $P$.

For $a=1$:
$1 * 1 = (1 \times 1 + 3) \pmod{8} = (1 + 3) \pmod{8} = 4 \pmod{8} = 4$
$1 * 3 = (1 \times 3 + 3) \pmod{8} = (3 + 3) \pmod{8} = 6 \pmod{8} = 6$
$1 * 5 = (1 \times 5 + 3) \pmod{8} = (5 + 3) \pmod{8} = 8 \pmod{8} = 0$
$1 * 6 = (1 \times 6 + 3) \pmod{8} = (6 + 3) \pmod{8} = 9 \pmod{8} = 1$

For $a=3$:
$3 * 1 = (3 \times 1 + 3) \pmod{8} = (3 + 3) \pmod{8} = 6 \pmod{8} = 6$
$3 * 3 = (3 \times 3 + 3) \pmod{8} = (9 + 3) \pmod{8} = 12 \pmod{8} = 4$
$3 * 5 = (3 \times 5 + 3) \pmod{8} = (15 + 3) \pmod{8} = 18 \pmod{8} = 2$
$3 * 6 = (3 \times 6 + 3) \pmod{8} = (18 + 3) \pmod{8} = 21 \pmod{8} = 5$

For $a=5$:
$5 * 1 = (5 \times 1 + 3) \pmod{8} = (5 + 3) \pmod{8} = 8 \pmod{8} = 0$
$5 * 3 = (5 \times 3 + 3) \pmod{8} = (15 + 3) \pmod{8} = 18 \pmod{8} = 2$
$5 * 5 = (5 \times 5 + 3) \pmod{8} = (25 + 3) \pmod{8} = 28 \pmod{8} = 4$
$5 * 6 = (5 \times 6 + 3) \pmod{8} = (30 + 3) \pmod{8} = 33 \pmod{8} = 1$

For $a=6$:
$6 * 1 = (6 \times 1 + 3) \pmod{8} = (6 + 3) \pmod{8} = 9 \pmod{8} = 1$
$6 * 3 = (6 \times 3 + 3) \pmod{8} = (18 + 3) \pmod{8} = 21 \pmod{8} = 5$
$6 * 5 = (6 \times 5 + 3) \pmod{8} = (30 + 3) \pmod{8} = 33 \pmod{8} = 1$
$6 * 6 = (6 \times 6 + 3) \pmod{8} = (36 + 3) \pmod{8} = 39 \pmod{8} = 7$

The completed table is:

The completed table is: