If $$p$$ and $$q$$ are the lengths of the perpendiculars from the origin on the lines, $$x\text{cosec}\alpha - y\sec\alpha = k\cot 2\alpha$$ and $$x\sin\alpha + y\cos\alpha = k\sin 2\alpha$$ respectively, then $$k^2$$ is equal to:

The first line is  $$x\csc\alpha - y\sec\alpha = k\cot 2\alpha.$$  We rewrite it in the standard straight-line form $$Ax+By+C=0$$ by bringing the right hand side to the left:

$$x\csc\alpha - y\sec\alpha - k\cot 2\alpha = 0.$$

Here $$A=\csc\alpha,\; B=-\sec\alpha,\; C=-k\cot 2\alpha.$$

For a line $$Ax+By+C=0,$$ the perpendicular distance of the origin $$(0,0)$$ from the line is given by the well-known formula  $$\displaystyle d=\frac{|C|}{\sqrt{A^{2}+B^{2}}}\;.$$  Stating this first keeps every step clear.

Applying the formula to the first line, the length of the perpendicular from the origin is

$$p=\frac{|\, -k\cot 2\alpha\, |}{\sqrt{\csc^{2}\alpha+(-\sec\alpha)^{2}}} =\frac{|k|\;|\cot 2\alpha|}{\sqrt{\csc^{2}\alpha+\sec^{2}\alpha}}.$$

Now we express the denominator completely in $$\sin\alpha$$ and $$\cos\alpha$$:

$$\csc^{2}\alpha=\frac{1}{\sin^{2}\alpha},\qquad \sec^{2}\alpha=\frac{1}{\cos^{2}\alpha}.$$

Therefore

$$\csc^{2}\alpha+\sec^{2}\alpha =\frac{1}{\sin^{2}\alpha}+\frac{1}{\cos^{2}\alpha} =\frac{\cos^{2}\alpha+\sin^{2}\alpha}{\sin^{2}\alpha\cos^{2}\alpha} =\frac{1}{\sin^{2}\alpha\cos^{2}\alpha}.$$

So

$$\sqrt{\csc^{2}\alpha+\sec^{2}\alpha} =\frac{1}{\sin\alpha\cos\alpha}.$$

Substituting this back, we obtain

$$p=|k|\;|\cot 2\alpha|\;(\sin\alpha\cos\alpha).$$

But $$\sin\alpha\cos\alpha=\dfrac{\sin 2\alpha}{2}$$ and also $$\cot 2\alpha=\dfrac{\cos 2\alpha}{\sin 2\alpha}.$$ Hence

$$p=|k|\;\Bigl|\frac{\cos 2\alpha}{\sin 2\alpha}\Bigr|\;\frac{\sin 2\alpha}{2} =\frac{|k|\;|\cos 2\alpha|}{2}.$$

Squaring to remove the absolute value,

$$p^{2}=\frac{k^{2}\cos^{2}2\alpha}{4}.\qquad(1)$$

Now we treat the second line $$x\sin\alpha+y\cos\alpha=k\sin 2\alpha.$$  Writing it as $$x\sin\alpha+y\cos\alpha-k\sin 2\alpha=0,$$ we have

$$A=\sin\alpha,\;B=\cos\alpha,\;C=-k\sin 2\alpha.$$

The distance formula gives

$$q=\frac{|\, -k\sin 2\alpha\, |}{\sqrt{\sin^{2}\alpha+\cos^{2}\alpha}} =\frac{|k|\;|\sin 2\alpha|}{\sqrt{1}} =|k|\;|\sin 2\alpha|.$$

Squaring,

$$q^{2}=k^{2}\sin^{2}2\alpha.\qquad(2)$$

We now possess two equations connecting $$p^{2},q^{2}$$ and $$k^{2}$$:

$$p^{2}=\dfrac{k^{2}\cos^{2}2\alpha}{4},\qquad q^{2}=k^{2}\sin^{2}2\alpha.$$

From the first, $$k^{2}=4p^{2}/\cos^{2}2\alpha,$$ and from the second, $$k^{2}=q^{2}/\sin^{2}2\alpha.$$ Equating these two expressions for $$k^{2}$$ (because they are the same quantity) we have

$$\frac{4p^{2}}{\cos^{2}2\alpha}=\frac{q^{2}}{\sin^{2}2\alpha}.$$

Cross-multiplying,

$$4p^{2}\sin^{2}2\alpha=q^{2}\cos^{2}2\alpha.$$

Bring the terms involving $$\sin^{2}2\alpha$$ to one side:

$$4p^{2}\sin^{2}2\alpha+q^{2}\sin^{2}2\alpha=q^{2}.$$

Factorising,

$$\sin^{2}2\alpha\,(4p^{2}+q^{2})=q^{2}.$$

Solve for $$\sin^{2}2\alpha$$:

$$\sin^{2}2\alpha=\frac{q^{2}}{4p^{2}+q^{2}}.$$

Finally, substitute this value into $$k^{2}=q^{2}/\sin^{2}2\alpha$$ obtained from equation (2):

$$k^{2}=\frac{q^{2}}{\dfrac{q^{2}}{4p^{2}+q^{2}}} =q^{2}\;\frac{4p^{2}+q^{2}}{q^{2}} =4p^{2}+q^{2}.$$

Thus we have reached the required result:

$$k^{2}=4p^{2}+q^{2}.$$

Hence, the correct answer is Option D.