`ABCD` is a parallelogram whose diagonals meet at P. If O is a fixed point, then `vec(OA)+vec(OB)+vec(OC)+vec(OD)` equals :

Options

A. (a) `vec(OP)`
B. (b) `2vec(OP)`
C. (c) `3vec(OP)`
D. (d) `4vec(OP)`

Explanation

To solve the problem, we need to find the expression for \( \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} \) in terms of the vector \( \vec{OP} \), where \( P \) is the midpoint of the diagonals of the parallelogram \( ABCD \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let \( O \) be a fixed point.
- Let \( A, B, C, D \) be the vertices of the parallelogram \( ABCD \).
- The diagonals \( AC \) and \( BD \) intersect at point \( P \).

2. **Using Vector Addition**:
- We can express the vectors from point \( O \) to each vertex of the parallelogram:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} \]

3. **Applying the Properties of Parallelograms**:
- Since \( P \) is the midpoint of both diagonals \( AC \) and \( BD \), we have:
\[ \vec{OP} = \frac{1}{2}(\vec{OA} + \vec{OC}) = \frac{1}{2}(\vec{OB} + \vec{OD}) \]

4. **Expressing \( \vec{OA} + \vec{OC} \) and \( \vec{OB} + \vec{OD} \)**:
- Rearranging gives us:
\[ \vec{OA} + \vec{OC} = 2\vec{OP} \]
\[ \vec{OB} + \vec{OD} = 2\vec{OP} \]

5. **Combining the Results**:
- Now, we can add these two results:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = (\vec{OA} + \vec{OC}) + (\vec{OB} + \vec{OD}) = 2\vec{OP} + 2\vec{OP} = 4\vec{OP} \]

6. **Final Result**:
- Therefore, we conclude that:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{OP} \]

### Conclusion:
The final answer is:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{OP} \]