= Solution
Use <degree elevation of Bernstein coefficients>. If $b_j$ are degree-$p$ <Bézier curve> controls, the identical curve has degree-$(p+1)$ controls
$$
b_0^+=b_0,\qquad b_{p+1}^+=b_p,\qquad b_j^+=\frac{j}{p+1}b_{j-1}+\left(1-\frac{j}{p+1}\right)b_j\quad(1\leq j\leq p).
$$
This follows by multiplying each degree-$p$ <Bernstein basis> polynomial by $(1-t)+t$ and collecting the degree-$(p+1)$ terms, so the operation changes the representation, not the curve.
For the quadratic controls in part (a), the cubic controls are
$$
D_0=B_0,\quad D_1=\frac{B_0+2B_1}{3}=\frac{P_{i-1}+5P_i}{6},\quad D_2=\frac{2B_1+B_2}{3}=\frac{5P_i+P_{i+1}}6,\quad D_3=B_2.
$$
Raise these cubic <Bézier curve> controls once more. Then
$$
E_0=B_0,\quad E_1=\tfrac14D_0+\tfrac34D_1=\tfrac12(B_0+B_1),\quad E_2=\tfrac12(D_1+D_2)=\tfrac16(B_0+4B_1+B_2),
$$
$$
E_3=\tfrac34D_2+\tfrac14D_3=\tfrac12(B_1+B_2),\qquad E_4=B_2.
$$
Thus the requested quartic <Bézier curve> controls, expressed in the original polygon, are
$$
\boxed{\begin{aligned}
E_0&=(P_{i-1}+P_i)/2,\\
E_1&=(P_{i-1}+3P_i)/4,\\
E_2&=(P_{i-1}+10P_i+P_{i+1})/12,\\
E_3&=(3P_i+P_{i+1})/4,\\
E_4&=(P_i+P_{i+1})/2.
\end{aligned}}
$$
Each expression is an <affine combination> whose coefficients sum to one. Endpoint positions and derivatives are preserved at each degree-raising step, as is every point of the span.
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