1. Let $P(x,y)$ be the variable point.

   Given fixed points: $O(0,0), \quad A(0,1)$

2. Write the three sides of $\triangle AOP$:

   • $OA = 1$
   • $OP = \sqrt{x^2+y^2}$
   • $AP = \sqrt{x^2+(y-1)^2}$

3. Use the perimeter condition:

   Perimeter of $\triangle AOP$ is $4$, so $OA+OP+AP=4$ $1+\sqrt{x^2+y^2}+\sqrt{x^2+(y-1)^2}=4$ $\sqrt{x^2+y^2}+\sqrt{x^2+(y-1)^2}=3$

4. This is the standard definition of an ellipse with foci at $O(0,0)$ and $A(0,1)$.

   So:

   • Sum of distances from the two foci $= 3 = 2a$
   • Hence, $a=\frac{3}{2}$

5. Find the center and focal distance:

   The center is midpoint of $(0,0)$ and $(0,1)$: $\left(0,\frac12\right)$

   Distance between foci is $1$, so $2c=1 \implies c=\frac12$

6. Find $b^2$ using $b^2=a^2-c^2$:

   $a^2=\frac{9}{4}, \quad c^2=\frac14$ $b^2=a^2-c^2=\frac94-\frac14=2$

7. Since the major axis is along the y-axis, the equation of ellipse is $\frac{x^2}{b^2}+\frac{\left(y-\frac12\right)^2}{a^2}=1$

   Substitute $a^2=\frac94$, $b^2=2$: $\frac{x^2}{2}+\frac{\left(y-\frac12\right)^2}{9/4}=1$

8. Simplify:

   $\frac{x^2}{2}+\frac{4}{9}\left(y-\frac12\right)^2=1$

   Multiply by $18$: $9x^2+8\left(y-\frac12\right)^2=18$

   Expand: $9x^2+8\left(y^2-y+\frac14\right)=18$ $9x^2+8y^2-8y+2=18$ $9x^2+8y^2-8y=16$

9. Match with options:

   $\boxed{9x^2+8y^2-8y=16}$

   This is Option A.

10. Comparison with stored answer:

Stored correct answer: A

Our derived answer: A

Hence, they agree.