You have reduced the problem to showing that ${\displaystyle \sum_{k=1}^n p_k^2 + \sum_{k=1}^n {1 \over p_k^2} \geq {1 \over n} + n^3}$. Note that equality is achieved here when each $p_k = {1 \over n}$, when the first sum adds to ${1 \over n}$ and the second sum adds to $n^3$. Thus it makes sense to try to show that the first sum is at least ${1 \over n}$ and the second sum is at least $n^3$.
For the first sum, one can use the Cauchy-Schwarz inequality to conclude that $n \sum_{k=1}^n {p_k}^2 \geq (\sum_{k=1}^n {p_k})^2 = 1$, so that the first sum is in fact at least ${1 \over n}$. For the second sum, one similarly has $n \sum_{k=1}^n {1 \over p_k^2} \geq (\sum_{k=1}^n {1 \over p_k})^2$ or equivalently, $\sum_{k=1}^n {1 \over p_k^2} \geq {1 \over n} (\sum_{k=1}^n {1 \over p_k})^2$. To analyze $\sum_{k=1}^n {1 \over p_k}$, one may use the arithmetic-harmonic inequality, which says that for positive real numbers $a_1,...,a_n$ one has ${a_1 + ... + a_n \over n} \geq {n \over {1 \over a_1} + ... + {1 \over a_n}}$ Applying this to $a_k = {1 \over p_k}$ gives that $\sum_{k=1}^n {1 \over p_k} \geq n^2{1 \over \sum_{k=1}^n p_k}$ $ = n^2$ So the second sum is at least ${1 \over n}(n^2)^2 = n^3$ as needed.
Edit: Naturally, in the time it took to write this out, someone already solved it.. such is life :)