Original Loading
$M = 50P ~ \text{N}\cdot\text{mm}$
Equivalent Loading
Top fiber (σ = 80 MPa compression)
$\sigma_{top} = -\sigma_a - \sigma_f$
$\sigma_{top} = -\dfrac{P}{A} - \dfrac{6M}{bd^2}$
$-80 = -\dfrac{P}{40(200)} - \dfrac{6(50P)}{40(200d^2)}$
$-80 = -\dfrac{P}{3,200}$
$P = 256,000 ~ \text{N} = 256 ~ \text{kN}$
Check the bottom fiber
$\sigma_{bottom} = -\sigma_a + \sigma_f$
$\sigma_{bottom} = -\dfrac{P}{A} + \dfrac{6M}{bd^2}$
$\sigma_{bottom} = -\dfrac{256(1000)}{40(200)} + \dfrac{6(50 \times 256)(1000)}{40(200^2)}$
$\sigma_{bottom} = 16 ~ \text{MPa tension} \lt 40 ~ \text{MPa}$ (okay)
Thus,
$P = 256 ~ \text{kN}$ answer