Calculate the PH ?

HBr is a strong acid. Each mole of HBr completely ionizes to give 1 mole of H₃O⁺.
[H₃O⁺] from HBr = [HBr]ₒ = 1.3 × 10⁻⁹ M
[H₃O⁺] form HBr is too small. Hence, the ionization of water must be considered.

                    2H₂O(ℓ)   ⇌     H₃O⁺(aq)    +    OH⁻      Kw = 1.0 × 10⁻¹⁴
Initial (M):        --                 1.3×10⁻⁹             0
Change (M):    --                      +y                +y
Eqm (M):          --             (1.3×10⁻⁹ + y)        y

At equilibrium:
Kw = [H₃O⁺] [OH⁻]
1.0 × 10⁻¹⁴ = (1.3 × 10⁻⁹ + y) y
y² + (1.3 × 10⁻⁹)y - (1.0 × 10⁻¹⁴) = 0
y = {-(1.3 × 10⁻⁹) ± √[(1.3 × 10⁻⁹)² + 4*(1.0 × 10⁻¹⁴)] / 2
y = 9.94 × 10⁻⁸  or  y = -1.01 × 10⁻⁷ (rejected)
[H₃O⁺] = (1.3 × 10⁻⁹ + 9.94 × 10⁻⁸) M = 1.007 × 10⁻⁷ M

pH = -log[H₃O⁺] = -log(1.007 × 10⁻⁷) = 7.00

This is a very common "trick" question related to pH. Normally, you would just say that because HBr is a strong acid, pH = -log (1.3X10^-9) = 8.89.  But this makes no sense because you have added a strong acid to water and ended up with a solution with a pH>7, which makes the solution basic.

What is happening here is that in most solutions, you ignore the autoionization of water that produces [H+] = 1X10^-7 M. We usually ignore this because we are usually dealing with solutions of acids and bases of much higher concentrations where this autoionization is relatively insignificant. But, when dealing with very low concentrations of acids and bases, this is no longer insignificant. 

So, in this solution, you can approximate
[H+] = 1.013X10^-7 M
pH = 6.99