The correct answer is $\frac{{{\text{WL}}}}{{4{\text{P}}}}$.
The central dip $h$ of a bent tendon is the vertical distance between the tendon and the horizontal line passing through the supports. The tendon is required to balance a concentrated load $W$ at the center of the span $L$.
The bending moment at any point on the tendon is given by $M = Wx$, where $x$ is the distance from the point to the center of the span. The maximum bending moment occurs at the center of the span, where $x = L/2$. Therefore, the maximum bending moment is $M = WL/2$.
The stress in the tendon is given by $\sigma = \frac{M}{I}$, where $I$ is the moment of inertia of the tendon. The moment of inertia of a bent tendon is given by $I = \frac{bh^3}{12}$, where $b$ is the width of the tendon and $h$ is the central dip.
The maximum stress in the tendon occurs at the center of the span, where $x = L/2$. Therefore, the maximum stress in the tendon is $\sigma = \frac{WL}{2} \cdot \frac{bh^3}{12}$.
The tendon is safe if the maximum stress in the tendon is less than the yield stress of the material. The yield stress of the material is denoted by $\sigma_y$. Therefore, the tendon is safe if $\frac{WL}{2} \cdot \frac{bh^3}{12} \leq \sigma_y$.
Solving for $h$, we get $h \geq \frac{2WL}{{3\sigma_y b}}$.
The load $W$ is given in the question. The yield stress $\sigma_y$ and the width $b$ of the tendon are known from the design of the tendon. Therefore, we can calculate the minimum central dip $h$ required to balance the concentrated load $W$ at the center of the span $L$.
The other options are incorrect because they do not satisfy the condition $\frac{WL}{2} \cdot \frac{bh^3}{12} \leq \sigma_y$.