= Solution
For a chosen coordinate and symbol $\varepsilon\in\{0,1\}$, the <shortened code> retains the words having that symbol in the chosen coordinate and then deletes that coordinate. For the last coordinate,
$$
\overline C_\varepsilon
=\{(x_1,\ldots,x_{n-1}):(x_1,\ldots,x_{n-1},\varepsilon)\in C\}.
$$
At least one of the two coordinate fibres contains at least $\lceil m/2\rceil$ words. Choose that fibre and, if necessary, discard surplus words. Because all retained words agree in the deleted coordinate, their mutual distances do not change. Hence one can always obtain
$$
\boxed{\overline C\text{ with guaranteed parameters }
[n-1,\lceil m/2\rceil,d]}.
$$
Without discarding words, its size is that of the chosen fibre and its minimum distance is at least $d$.
For the final calculation, the given code is the whole space $\mathbb F_2^3$, so its parity extension is the length-four <even-weight binary code>. In a <binary symmetric channel>, parity fails to notice a nonzero error exactly when an even number of bits flips. The possible error weights are therefore two and four. Their total probability is
$$
\boxed{
\binom42p^2(1-p)^2+p^4
=6p^2(1-p)^2+p^4
}.
$$
This excludes the weight-zero event, since it is not an <undetected error>.
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