For what value of k the expression p + 1/4 + p + k^2 is perfect square?

Answer: Option A

$$\begin{array}{rl}& p+\frac{1}{4}+\sqrt{p}+k^2\\ & =p+\sqrt{p}+\left(k^2+\frac{1}{4}\right)\\ & ={\left(\sqrt{p}\right)}^2+2\times \frac{1}{2}\times \sqrt{p}+\left(k^2+\frac{1}{4}\right)\\ & ={\text{A}}^2+\text{2}\times \text{A}\times \text{B}+{\text{B}}^2\\ & \text{A}=\sqrt{p}\\ & {\text{B}}^2=\left(k^2+\frac{1}{4}\right)\\ & \text{B}=\frac{1}{2}\\ & \therefore k^2+\frac{1}{4}={\left(\frac{1}{2}\right)}^2\\ & \Rightarrow k^2+\frac{1}{4}=\frac{1}{4}\\ & \Rightarrow k^2=0\\ & \Rightarrow k=0\end{array}$$