= Solution
For any finite-dimensional bounded <cochain complex> $(V^*,d)$ over a <field>, let $r_q=\operatorname{rank}(d:V^q\to V^{q+1})$, with zero ranks outside its degree range. Then
$$
\dim H^q(V,d)=\dim V^q-r_q-r_{q-1}.
$$
Taking the alternating sum cancels the two rank sums. Thus taking <cohomology> preserves the <Euler characteristic> of a finite graded complex. Apply this to $V^q=H^q(M;\mathbb F_p)$ and $d=\beta$, using part (b).
A closed three-dimensional <manifold> has finite-dimensional <cohomology>, vanishing above degree three. Its <Euler characteristic> is zero, even when it is not orientable: <Poincare duality> with $\mathbb F_2$ coefficients gives $b_q=b_{3-q}$, so the alternating sum vanishes. A finite triangulation, or finite <CW complex> model, shows that the alternating sum is the same integer with coefficients in any field, since it equals the alternating count of cells. Therefore
$$
\boxed{\chi\bigl(H\beta^*(M;p)\bigr)=\chi(M)=0.}
$$
The primeness assumption makes $\mathbb Z/p$ a field; no orientation assumption on $M$ is needed.
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