= Solution
Under ideal independent <mixing on crystallographic sites>, a <mineral endmember>'s <thermodynamic activity> is the product of its required site fractions, one factor per mixed site. <Muscovite> has one K and three Al per formula unit, <sanidine> has one K and one Al, and <corundum> has two Al. Interpreting the given Al fractions as the fractions on every relevant Al/Fe site gives
$$
a_{\rm mus}=0.95(0.9)^3=0.69255,\quad a_{\rm san}=0.75(0.9)=0.675,\quad a_{\rm cor}=(0.8)^2=0.64.
$$
Thus $A_s=a_{\rm san}a_{\rm cor}/a_{\rm mus}=0.6237816764$. The cubic power for <muscovite> includes its tetrahedral Al site and two octahedral Al sites; if site-specific fractions differed, bulk Al/(Al+Fe) alone would not determine this activity. The calculation uses the uniform-site ideal-mixing interpretation intended by the given data.
At $P=5000\ \mathrm{bar}$ with ideal water, solve
$$
T=\frac{98157-0.61(5000-1)}{164-8.3145\ln(5000A_s)},
$$
giving \b[$T_{\rm eq}=979.40\ \mathrm K=706.25\ ^\circ\mathrm C$]. Using the bulk <mole fraction> only once for a multisite <mineral endmember> would omit part of its ideal mixing <chemical potential>.
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