pH and pKa question on titration curve please????!!!?

It is not "equivalence point", but is "half equivalent point".

Consider a weak acid HA with dissociation constant Kₐ.
At the half equivalence point, a half of HA is neutralized to form H⁺ and A⁻.
Therefore, at this point, not considering the dissociation of HA, [HA]ₒ = [A⁻]ₒ.

Consider the dissociation of HA at the half equivalent point.
HA(aq) + H₂O(l) ⇌ A⁻(aq) + H₃O⁺(aq)
The dissociation is to a negligible extent, due to the small Kₐ and the common ion effect as the presence of A⁻ ions.
Therefore, [HA] ≈ [HA]ₒ and [A⁻] = [HA]ₒ
and thus [HA] ≈ [A⁻]

At equilibrium :
Kₐ = [A⁻] [H₃O⁺] / [HA]
Kₐ = [H₃O⁺] × ([A⁻]/[HA])

Taken negative on the both sides :
-log(Kₐ) = -log{[H₃O⁺] × ([A⁻]/[HA])}
-log(Kₐ) = -log[H₃O⁺] - log([A⁻]/[HA])

As -log(Kₐ) = pKₐ, -log[H₃O⁺] = pH and [HA] ≈ [A⁻] :
pKₐ = pH - log(1)
pKₐ = pH