Maths problem, how to do thanks?

Solution (REVISED) :
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∵ AB=AP, ∴△BAP is an isosceles △
=> ∠ABP=∠APB
Let ∠ABP=∠APB = b

From △ABQ,
∠AQB=90⁰ (∠ in semicircle)
∴ ∠BAQ = 90⁰ - b . . . . . . . . . . . . . ①
From △PQN,
∠QNP=90⁰ (Given that QN丄AP)
∴ ∠PQN = 90⁰ - b . . . . . . . . . . . . . ②
①&② : ∠BAQ = ∠PQN . . . . . . . . . . . . . (#)

Now, from △ABQ, ∠ABQ = 90⁰ - ∠BAQ. . . . . . . . . . . ③
Also, from pt. Q, ∠AQN = 90⁰ - ∠PQN. . . . . . . . . . . . .④
Concluding from (#), ③&④ : ∠ABQ = ∠AQN
Conclusion :
QN is tangent to the circle at Q (converse of ∠ in alternate segment)