Assuming P = A + B, then P ~ N(3.5, 16).

$$\text{VaR}(\alpha\%) = [-\mu_{_T} + \sigma_{_T} \times Z_\alpha] \\ = [-3.5 + 4 \times 1.645] \\ = 3.08$$

Note. The sum of two independent normally distributed random variables is normal. Its mean is the sum of the two means, and its variance is the sum of the two variances

Where $\sigma_{_T} = \sqrt{16} = 4$