The quadrilateral $ABCD$ is inscribed in the parabola $y=x^2$. It is known that angle $BAD=90$, the dioganal $AC$ is parallel to the axis $Ox$ and $AC$ is the bisector of the angle BAD.
Find the area of the quadrilateral $ABCD$ if the length of the dioganal $BD$ is equal to $p$.
1. **Identify the coordinates of points \(A\) and \(C\)**:
   Since \(A\) and \(C\) are on the parabola \(y = x^2\) and \(AC\) is parallel to the \(x\)-axis, we can write:
   \[
   A = (a, a^2) \quad \text{and} \quad C = (-a, a^2)
   \]
   Here, \(a > 0\).

2. **Determine the equations of lines \(AB\) and \(AD\)**:
   Given that \(\angle BAD = 90^\circ\) and \(AC\) is the angle bisector of \(\angle BAD\), we can find the slopes of \(AB\) and \(AD\):
   - The slope of \(AB\) is \(-1\) (since \(AC\) bisects \(\angle BAD\) and \(\angle BAC = 45^\circ\)).
   - The slope of \(AD\) is \(1\).

   Using the point-slope form of the line equation:
   \[
   AB: y - a^2 = -1(x - a) \implies y = -x + a^2 + a
   \]
   \[
   AD: y - a^2 = 1(x - a) \implies y = x + a^2 - a
   \]

3. **Find the coordinates of points \(B\) and \(D\)**:
   - For \(B\), substitute \(x = -a - 1\) into the equation of the parabola:
     \[
     B = (-a - 1, (-a - 1)^2) = (-a - 1, a^2 + 2a + 1)
     \]
   - For \(D\), substitute \(x = 1 - a\) into the equation of the parabola:
     \[
     D = (1 - a, (1 - a)^2) = (1 - a, 1 - 2a + a^2)
     \]

4. **Verify the condition \(D_y < A_y\)**:
   \[
   D_y = 1 - 2a + a^2 < a^2 \implies 1 - 2a < 0 \implies a > \frac{1}{2}
   \]

5. **Calculate the length of diagonal \(BD\)**:
   \[
   BD^2 = (x_D - x_B)^2 + (y_D - y_B)^2
   \]
   \[
   x_D - x_B = (1 - a) - (-a - 1) = 2 - 2a
   \]
   \[
   y_D - y_B = (1 - 2a + a^2) - (a^2 + 2a + 1) = -4a
   \]
   \[
   BD^2 = (2 - 2a)^2 + (-4a)^2 = 4(1 - a)^2 + 16a^2 = 4(1 - 2a + a^2) + 16a^2 = 4 + 16a^2
   \]
   Given \(BD = p\):
   \[
   p^2 = 4 + 16a^2 \implies a^2 = \frac{p^2 - 4}{16}
   \]

6. **Calculate the area of quadrilateral \(ABCD\)**:
   The area of \(ABCD\) can be found by summing the areas of triangles \(ABC\) and \(ADC\):
   \[
   S_{ABCD} = S_{ABC} + S_{ADC}
   \]
   Since \(AC\) is parallel to the \(x\)-axis and the height difference is the same for both triangles:
   \[
   S_{ABC} = \frac{1}{2} \times AC \times |B_y - A_y| = \frac{1}{2} \times 2a \times (a^2 + 2a + 1 - a^2) = \frac{1}{2} \times 2a \times (2a + 1) = a(2a + 1)
   \]
   \[
   S_{ADC} = \frac{1}{2} \times AC \times |D_y - A_y| = \frac{1}{2} \times 2a \times (a^2 - 2a + 1 - a^2) = \frac{1}{2} \times 2a \times (1 - 2a) = a(1 - 2a)
   \]
   \[
   S_{ABCD} = a(2a + 1) + a(1 - 2a) = a \times 2a + a \times 1 + a \times 1 - a \times 2a = 4a^2
   \]
   Substituting \(a^2 = \frac{p^2 - 4}{16}\):
   \[
   S_{ABCD} = 4 \times \frac{p^2 - 4}{16} = \frac{p^2 - 4}{4}
   \]

The final answer is \(\boxed{\frac{p^2 - 4}{4}}\).