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For a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations is its number of conjugacy classes, and the sum of squares of irreducible representation dimensions equals the group order. There are therefore five irreducibles for $S_4$, with squared dimensions summing to $24$.

First establish that part (c) found every one-dimensional representation. In such a representation, conjugate elements have the same scalar image. All transpositions are conjugate and square to the identity, so they all have one common image $s$ with $s^2=1$. Transpositions generate $S_4$, hence $s=1$ gives exactly the trivial representation, while $s=-1$ gives exactly the sign representation. There are precisely two degree-one irreducibles.

The other three dimensions $d_1,d_2,d_3$ are integers at least two and satisfyNo one of them can be at least four, since $4^2+2^2+2^2=24>22$. They are therefore all two or three. Starting from three twos gives sum of squares $12$; each replacement of a two by a three adds $5$. To reach $22$ requires exactly two replacements. Thus the complete list isThe count of conjugacy classes guarantees that these are all the inequivalent complex irreducible representations, not just dimensions of a proposed incomplete list.

Write $n=q(p-1)+r$, $0\le r<p-1$. The regular partition labelling the sign representation in characteristic $p>0$ is $\lambda =((q+1{)}^r,q^{p-1-r})$, with zero parts removed. Its conjugate partition is $((p-1{)}^q,r)$, which satisfies the invariant vector criterion for a Specht module. The sign-twisted duality of Specht modules turns this invariant line into a sign quotient of $S^{\lambda }$, necessarily its unique simple head. Conversely, regularity says consecutive column heights differ by less than $p$; combining this with the invariant congruences forces these heights to be $p-1$ except possibly the last. For $p=2$, the label is $(n)$ because sign equals the trivial representation.