To solve for the tensions in the strings, we resolve the forces acting on P horizontally and vertically. Let the tension in PA be T_A and in PB be T_B.

Using the geometry of the triangle formed by the strings and the vertical line, we have:

\(T_A \times \frac{40}{50} + T_B \times \frac{40}{104} = 21\)

Solving for T_A and T_B:

\(T_A \times \frac{30}{50} = T_B \times \frac{96}{104}\)

From the first equation, we find:

\(T_A = 20 \text{ N}\)

Substituting T_A into the second equation gives:

\(T_B = 13 \text{ N}\)

Thus, the tension in AP is 20 N and the tension in BP is 13 N.