= Solution
Code each factor level by $x_j\in\{-1,1\}$. A nonconstant <factorial contrast> indexed by a nonempty factor set $A$ has sign column $g_A(x)=\prod_{j\in A}x_j$. Flipping any coordinate in $A$ reverses that sign and pairs the <treatment> combinations, so exactly $2^{m-1}$ have each sign. Allocating the two signs to two <experimental blocks> therefore divides the full <factorial design> into equal <experimental blocks>. The <contrast> is confounded with <experimental blocks> because its column is constant within each <experimental block>: changing its coefficient and compensating by the two <experimental block> <intercepts> leaves every fitted value unchanged. It has no information separate from <experimental block> differences. Randomize the allocation of the two groups to days and the run order within days.
Write $A,B,C$ for temperature, pressure and catalyst, respectively, with low levels and catalyst one coded minus. In the supplied allocation, $ABC=-1$ on day one and $ABC=1$ on day two. Thus \b[the three-factor <interaction> $ABC$ is confounded with days]. Every other nonconstant factorial column is <orthogonal> to the day <contrast>, because its product with $ABC$ is a nonconstant sign column on the full cube and consequently sums to zero.
If all <interactions> are negligible, the fitted model contains an <intercept>, one day <contrast> and the three <main effects>. The corrected total <statistical degrees of freedom> partition as
$$
\boxed{7=1\ \text{(days)}+1\ \text{(temperature)}+1\ \text{(pressure)}
+1\ \text{(catalyst)}+3\ \text{(residual)}.}
$$
The three residual <statistical degrees of freedom> are the unused <two-factor interaction> columns $AB,AC,BC$, pooled as error under the stated negligible-interaction assumption. There is no independent pure-error replication of <treatment> combinations.
For four days with two runs each, \b[all three <main effects> can be estimated]. One regular allocation pairs each <treatment> combination with its complete opposite:
$$
\begin{array}{c|cc}
\text{block}&\text{first }(A,B,C)&\text{second }(A,B,C)\\\hline
1&(-,-,-)&(+,+,+)\\
2&(-,-,+)&(+,+,-)\\
3&(-,+,-)&(+,-,+)\\
4&(-,+,+)&(+,-,-)
\end{array}
$$
The block-generating columns can be $AB$ and $AC$; their product $BC$ is also constant in <experimental blocks>. Thus $AB,AC,BC$ span the three <experimental block> <contrasts>, while $A,B,C$ and $ABC$ vary within <experimental blocks>. The <main effects> remain mutually <orthogonal> and separate from the <experimental block> effects; with an additive main-effect model there is one residual degree of freedom, represented by $ABC$.
For the further request about $AC$, the distinction between regular confounding and unrestricted pairing is important. \b[A regular four-block allocation cannot retain all three <main effects> and $AC$.] Its three confounded columns form a rank-two defining subgroup. To preserve <main effects>, these columns must lie in $\{AB,AC,BC,ABC\}$. If the subgroup includes $ABC$ and a two-factor column, their product is a main-effect column, which is forbidden. Hence the only permissible nonidentity subgroup is $\{AB,AC,BC\}$, and it necessarily confounds $AC$. This proves impossibility in the usual regular factorial-blocking framework, rather than relying on a count of available <statistical degrees of freedom>.
The literal request does not explicitly restrict arrangements to regular <experimental blocks>. If arbitrary pairings are permitted, a nonregular construction can estimate $A,B,C,AC$ together, assuming the other <interactions> are omitted from the model. An explicit allocation is
$$
\begin{array}{c|cc}
\text{block}&\text{first }(A,B,C)&\text{second }(A,B,C)\\\hline
1&(-,-,-)&(-,-,+)\\
2&(-,+,-)&(+,-,-)\\
3&(-,+,+)&(+,+,+)\\
4&(+,-,+)&(+,+,-)
\end{array}
$$
Every <treatment> combination occurs exactly once. Fit one unrestricted <intercept> per day and the four coefficients $\beta_A,\beta_B,\beta_C,\beta_{AC}$. By <estimability from within-block differences>, subtracting the second run from the first in each <experimental block> removes all day effects and yields the coefficient <matrix>
$$
D=\begin{pmatrix}
0&0&-2&2\\
-2&2&0&2\\
-2&0&0&-2\\
0&-2&2&2
\end{pmatrix},\qquad\det D=-64\ne0.
$$
Thus all four coefficients are uniquely determined from the four day differences. In this broader design class the answer is \b[yes, under the reduced four-effect model]. The resulting fit is saturated: four <experimental block> <intercepts> plus four <treatment> coefficients use all eight observations, leaving no residual estimate of error <variance>. Replication or external error information would be needed for inference. This does not contradict the impossibility for regular <orthogonal> confounding, and it does not claim simultaneous unbiased estimation when arbitrary additional <interactions> are retained.
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