This question is borderline for appropriateness here, so I'll sketch how to proceed and omit the details. Let $f(p) = p(1-p)^t$. Notice that
$$f''(p) = (1-p)^{t-2} (t(t-1) p + 2t (1-p))= t (1-p)^{t-2} ((1+t)p - 2) $$
So $f$ is concave on $[0,2/(1+t)]$ and convex on $[2/(1+t), 1]$.

Take any $(p_1, \ldots, p_n)$. If two of the $p_i$ are in the interval $(2/(1+t), 1)$, then push them apart while maintaining their sum until one of them hits the boundary of the interval. This will increase $f(p_i)$, as $f$ is convex on this interval.

Since their sum is $\leq 1$, the smaller $p_i$ will hit $2/(1+t)$ before the larger one hits $1$. Repeating this argument, we can continue to push $p_i$'s apart until at most one $p_i$ is in the interval $(2/(1+t), 1)$; call it $v$.

Now, take all the other $p_i$ besides $v$ and replace them all by their common average. Since $f$ is concave on $[0,2/(1+t)]$, this will increase $\sum f(p_i)$. (This is Jensen's inequality.) So we have reduced to one of two cases: Either there is one $p_i$, called $v$ in $[2/(1+t), 1]$, and all the others are equal to $(1-v)/(n-1)$, or else all the $p_i$ are in $[0,2/(1+t)]$, and they have value $1/n$.

You now have a single variable function to optimize, and also one other value to compare it to. I haven't done the work, but it should be tractable from here.

I learned this approach from Kiran Kedlaya's notes on inequalities. It's really a shame that there is no course in the standard curriculum which teaches this.