Correct Answer: $\frac{20}{3}\ \text{cm}$

Solution :
Join OP and OT.
Let OT intersect PQ at a point R.
Then, TP = TQ (The lengths of tangents drawn from an external point to a circle are equal) and $\angle$PTR = $\angle$QTR.
⇒ TR $\perp$ PQ and TR bisect PQ.
⇒ PR = RQ = 4 cm
OP = 5 cm
So, OR ² = OP ² – PR ² = $\sqrt{5^2-4^2}=\sqrt{9}=3$ cm
Let TP = $x$ cm
and TR = $y$ cm
From right ΔTRP, we get,
TP ² = TR ² + PR ²
⇒ $x$ ² = $y$ ² + 16
⇒ $x$ ² − $y$ ² = 16-------------------------(i)
From right ΔOPT, we get
TP ² + OP ² = OT ²
⇒ $x$ ² + 5 ² = ($y$ + 3) ² [$\because$ OT ² = (OR + RT) ² ]
⇒ $x$ ² − $y$ ² = 6$y$ − 16------------------(ii)
From (i) and (ii), we get,
6$y$ − 16 = 16
⇒ 6$y$ = 32
⇒ $y$ = $\frac{16}{3}$
Putting the value of $y$ in equation (i), we get,
$x$ ² $=16 + (\frac{16}{3})^2=16+\frac{256}{9}=\frac{400}{9}=\frac{20}{3}\ \text{cm}$
Hence, the correct answer is $\frac{20}{3}\ \text{cm}$.