$$t_{100\%}$$ is the time required for the 100% completion of the reaction while $$t_{1/2}$$ is the time required for 50% of the reaction to be completed. Which of the following option correctly represents the relation between $$t_{100\%}$$ and $$t_{1/2}$$ for zero and first order reactions respectively?

For each order of reaction we start with its integrated rate law and then evaluate the times corresponding to 50 % and to 100 % conversion.

Case 1: Zero-order reaction
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Integrated rate law: 
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$$[A]=[A]_0-k\,t$$

Half-life
($$t_{1/2}$$): at 50 % conversion, $$[A]=\frac{[A]_0}{2}$$.
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Hence
$$\frac{[A]_0}{2}=[A]_0-k\,t_{1/2}\; \Longrightarrow\; t_{1/2}=\frac{[A]_0}{2k}$$
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Complete conversion ($$t_{100\%}$$): at 100 % conversion, $$[A]=0$$. 

Hence $$0=[A]_0-k\,t_{100\%}\; \Longrightarrow\; t_{100\%}=\frac{[A]_0}{k}$$

Taking the ratio,
$$\frac{t_{100\%}}{t_{1/2}}=\frac{\frac{[A]_0}{k}}{\frac{[A]_0}{2k}}=2$$
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Therefore for a zero-order reaction $$t_{100\%}=2\,t_{1/2}$$.

Case 2: First-order reaction
$$\text{}$$Integrated rate law: $$\ln\!\left(\frac{[A]_0}{[A]}\right)=k\,t$$

Half-life($$t_{1/2}$$): set $$[A]=\dfrac{[A]_0}{2}$$.
$$\ln 2 = k\,t_{1/2}\; \Longrightarrow\; t_{1/2}=\frac{\ln 2}{k}$$

Complete conversion ($$t_{100\%}$$): to reach 100 % conversion we would need $$[A]\rightarrow 0$$.
$$\lim_{[A]\to 0}\ln\!\left(\frac{[A]_0}{[A]}\right)=\infty$$, so $$t_{100\%}=\infty$$.
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Since any positive finite number raised to the power $$\infty$$ tends to $$\infty$$, we may symbolically write
$$t_{100\%}=(t_{1/2})^{\infty}$$ for a first-order reaction.

Combining both cases:
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Zero order $$t_{100\%}=2\,t_{1/2}$$
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First order $$t_{100\%}=(t_{1/2})^{\infty}$$
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Hence the correct choice is:
Option B which states $$t_{100\%}=2t_{1/2}$$ (zero order) and $$t_{100\%}=(t_{1/2})^{\infty}$$ (first order).