= Solution
The single secondary turn links the primary <magnetic flux> $\mu_0n\pi a^2I_p$, so its <mutual inductance> with the primary is
$$
M=\mu_0n\pi a^2=\frac{L}{nD}.
$$
Choose positive winding directions so that positive secondary <electric current> would reinforce the primary <magnetic flux>. <Faraday's law> and <Kirchhoff's voltage law> then give
$$
L_s\frac{dI_s}{dt}+RI_s=-M\frac{dI_p}{dt}=\omega MI_0\sin\omega t.
$$
After the transient proportional to $e^{-Rt/L_s}$ has decayed, put $\alpha=\tan^{-1}(\omega L_s/R)$. The steady <electric current> is
$$
\boxed{I_s(t)=\frac{\omega MI_0}{\sqrt{R^2+(\omega L_s)^2}}\sin(\omega t-\alpha).}
$$
Its <wave amplitude> is $\omega MI_0/\sqrt{R^2+(\omega L_s)^2}$ and it lags the induced <electromotive force> by $\alpha$. Relative to the primary $I_0\cos\omega t$, its <phase shift> is $-\pi/2-\alpha$, modulo $2\pi$. Reversing the chosen secondary winding direction adds $\pi$ to this signed <phase shift> and leaves all <power> predictions unchanged.
The lamp's mean <Joule heating> is
$$
\boxed{\langle P_R\rangle=\frac12R|I_s|^2=\frac{R\omega^2M^2I_0^2}{2[R^2+(\omega L_s)^2]}.}
$$
For $\omega L_s\ll R$ the response is nearly in phase with the induced <electromotive force>; for $\omega L_s\gg R$ the secondary <electric current> nearly opposes the primary <electric current> in the stated winding convention.
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