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First, we consider a particular line, $y=x-1,$ which passes through $F.$ Substituting, we get $$\frac{x^2}{2}+(x-1{)}^2=1.$$This simplifies to $3x^2-4x=x(3x-4)=0,$ so $x=0$ or $x=\frac{4}{3}.$ Thus, we can let $A=\left(\frac{4}{3},\frac{1}{3}\right)$ and $B=(0,-1).$

The slope of line $AP$ is then $\frac{1/3}{4/3-p}=\frac{1}{4-3p},$ and the slope of line $BP$ is $\frac{-1}{-p}=\frac{1}{p}.$ Since $\angle APF=\angle BPF,$ these slopes are negatives of each other, so $$\frac{1}{3p-4}=\frac{1}{p}.$$Then $p=3p-4,$ so $p=\boxed{2}.$

For a complete solution, we prove that this works for all chords $\overline{AB}$ that pass through $F.$ Let $A=(x_a,y_a)$ and $B=(x_b,y_b).$ Then the condition $\angle APF=\angle BPF$ is equivalent to $$\frac{y_a}{x_a-2}+\frac{y_b}{x_b-2}=0,$$or $y_a(x_b-2)+y_b(x_a-2)=0.$ Then $y_a{x}_b-2y_a+y_b{x}_a-2y_b=0.$

Let $y=m(x-1)$ be the equation of line $AB.$ Substituting, we get $$\frac{x^2}{2}+m^2(x-1{)}^2=1.$$This simplifies to $(2m^2+1)x^2-4m^{2}x+2m^2-2=0.$ By Vieta's formulas, $$x_a+x_b=\frac{4m^2}{2m^2+1}\text{and}{x}_a{x}_b=\frac{2m^2-2}{2m^2+1}.$$Then $$\begin{array}{rl}{y}_a{x}_b-2y_a+y_b{x}_a-2y_b& =m(x_a-1)x_b-2m(x_a-1)+m(x_b-1)x_a-2m(x_b-1)\\ & =2m{x}_a{x}_b-3m(x_a+x_b)+4m\\ & =2m\cdot \frac{2m^2-2}{2m^2+1}-3m\cdot \frac{4m^2}{2m^2+1}+4m\\ & =0.\end{array}$$Thus, $\angle APF=\angle BPF$ for all chords $\overline{AB}$ that pass through $F.$