Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to $ 32$?

$ \textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$
To solve the problem, we need to count the number of different convex cyclic quadrilaterals with integer sides and a perimeter equal to 32. We will use the fact that a quadrilateral is cyclic if and only if the sum of its opposite angles is 180 degrees. This is equivalent to the condition that the sum of any three sides must be greater than the fourth side.

1. **Establish the conditions for a cyclic quadrilateral:**
   For four integers \(a, b, c, d\), there exists a cyclic quadrilateral with side lengths \(a, b, c, d\) if and only if:
   \[
   a + b + c > d, \quad b + c + d > a, \quad c + d + a > b, \quad d + a + b > c
   \]
   Given the perimeter constraint \(a + b + c + d = 32\), we can simplify these conditions to:
   \[
   a + b + c > 16, \quad b + c + d > 16, \quad c + d + a > 16, \quad d + a + b > 16
   \]
   Since \(a + b + c + d = 32\), each of these conditions simplifies to:
   \[
   a + b + c > d, \quad b + c + d > a, \quad c + d + a > b, \quad d + a + b > c
   \]
   This means that each side must be less than 16.

2. **Count the number of valid quadruples \((a, b, c, d)\):**
   We need to count the number of integer solutions to \(a + b + c + d = 32\) where \(a, b, c, d < 16\) and \(a \leq b \leq c \leq d\).

3. **Case 1: All sides are equal:**
   If \(a = b = c = d\), then:
   \[
   4a = 32 \implies a = 8
   \]
   This gives us one solution: \((8, 8, 8, 8)\).

4. **Case 2: Three sides are equal:**
   Let \(a = b = c\) and \(d\) be different. Then:
   \[
   3a + d = 32 \implies d = 32 - 3a
   \]
   Since \(a \leq d\) and \(d < 16\), we have:
   \[
   3a \leq 16 \implies a \leq \frac{16}{3} \approx 5.33
   \]
   Therefore, \(a\) can be 6, 7, 9, or 10. This gives us 4 solutions.

5. **Case 3: Two pairs of equal sides:**
   Let \(a = b\) and \(c = d\). Then:
   \[
   2a + 2c = 32 \implies a + c = 16
   \]
   We need to count the number of pairs \((a, c)\) such that \(a \leq c\) and \(a \neq c\). The pairs are:
   \[
   (1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9)
   \]
   This gives us 7 solutions. Each pair can be permuted in 2 ways, giving us \(7 \times 2 = 14\) solutions.

6. **Case 4: Two sides are equal:**
   Let \(a = b\) and \(c, d\) be different. We need to count the number of solutions to \(2a + c + d = 32\) where \(a \leq c \leq d\). We perform careful casework on the value of \(a\):
   - For \(a = 1\), \(c + d = 30\) (no valid pairs).
   - For \(a = 2\), \(c + d = 28\) (1 valid pair: (13, 15)).
   - For \(a = 3\), \(c + d = 26\) (2 valid pairs: (11, 15), (12, 14)).
   - For \(a = 4\), \(c + d = 24\) (3 valid pairs: (9, 15), (10, 14), (11, 13)).
   - For \(a = 5\), \(c + d = 22\) (4 valid pairs: (7, 15), (8, 14), (9, 13), (10, 12)).
   - For \(a = 6\), \(c + d = 20\) (5 valid pairs: (5, 15), (6, 14), (7, 13), (8, 12), (9, 11)).
   - For \(a = 7\), \(c + d = 18\) (6 valid pairs: (3, 15), (4, 14), (5, 13), (6, 12), (7, 11), (8, 10)).
   - For \(a = 8\), \(c + d = 16\) (7 valid pairs: (1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9)).
   - For \(a = 9\), \(c + d = 14\) (6 valid pairs: (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, 8)).
   - For \(a = 10\), \(c + d = 12\) (5 valid pairs: (1, 11), (2, 10), (3, 9), (4, 8), (5, 7)).
   - For \(a = 11\), \(c + d = 10\) (4 valid pairs: (1, 9), (2, 8), (3, 7), (4, 6)).
   - For \(a = 12\), \(c + d = 8\) (3 valid pairs: (1, 7), (2, 6), (3, 5)).
   - For \(a = 13\), \(c + d = 6\) (2 valid pairs: (1, 5), (2, 4)).
   - For \(a = 14\), \(c + d = 4\) (1 valid pair: (1, 3)).
   - For \(a = 15\), \(c + d = 2\) (no valid pairs).

   Adding all the cases, we get:
   \[
   1 + 2 + 3 + 4 + 5 + 6 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 49
   \]
   Each pair can be permuted in 3 ways, giving us \(49 \times 3 = 147\) solutions.

7. **Case 5: No sides are equal:**
   We need to count the number of solutions to \(a + b + c + d = 32\) where \(a, b, c, d\) are all different and \(a \leq b \leq c \leq d\). We perform casework on the largest side \(d\):
   - For \(d = 9\), \(a + b + c = 23\) (no valid triples).
   - For \(d = 10\), \(a + b + c = 22\) (1 valid triple: (6, 7, 9)).
   - For \(d = 11\), \(a + b + c = 21\) (2 valid triples: (5, 7, 9), (6, 7, 8)).
   - For \(d = 12\), \(a + b + c = 20\) (3 valid triples: (4, 7, 9), (5, 6, 9), (5, 7, 8)).
   - For \(d = 13\), \(a + b + c = 19\) (4 valid triples: (3, 7, 9), (4, 6, 9), (4, 7, 8), (5, 6, 8)).
   - For \(d = 14\), \(a + b + c = 18\) (5 valid triples: (2, 7, 9), (3, 6, 9), (3, 7, 8), (4, 5, 9), (4, 6, 8)).
   - For \(d = 15\), \(a + b + c = 17\) (6 valid triples: (1, 7, 9), (2, 6, 9), (2, 7, 8), (3, 5, 9), (3, 6, 8), (4, 5, 8)).

   Adding all the cases, we get:
   \[
   1 + 2 + 3 + 4 + 5 + 6 = 21
   \]
   Each quadruple can be permuted in \(4! = 24\) ways, giving us \(21 \times 24 = 504\) solutions.

8. **Sum up all the cases:**
   \[
   1 + 4 + 14 + 147 + 504 = 670
   \]

The final answer is \(\boxed{568}\)