(a) Let $p_1\le \cdots\le p_n$ without loss of generality. Then I claim $\textstyle \sum_{i\le n}x_i^2\le p_1^{-2}$. Indeed, we have $1\ge p_1\textstyle \sum_{i\le n}x_i\implies p_1^{-2}\ge \textstyle\sum_{i\le n}x_i^2$. The equality is achieved for $(1/p_1,0,\dots,0)$. (b) Let $\|p\|^2 \triangleq \textstyle \sum_{i\le n}p_i^2$. Then I claim $\textstyle \sum_{i\le n}x_i^2 \ge \|p\|^{-2}$. Indeed, by Cauchy-Schwarz we have $\|p\|^2\cdot \textstyle \sum_{i\le n}x_i^2\ge \left(\textstyle \sum_{i\le n}x_ip_i\right)^2=1$, yielding the conclusion. The equality holds when $x_i=p_i/\|p\|^2$.