Write the hyperbola in standard form

Given $\frac{x^2}{\cos^2\alpha}-\frac{y^2}{\sin^2\alpha}=1$

Compare with the standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

So, $a^2=\cos^2\alpha, \qquad b^2=\sin^2\alpha.$

Check each quantity

Option A: Abscissae of vertices

For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ vertices are at $(\pm a,0).$ Hence their abscissae are $\pm a=\pm \cos\alpha.$

Since $\cos\alpha$ changes with $\alpha$, this is not constant.

So, A is false.

Option B: Abscissae of foci

For the hyperbola, $c^2=a^2+b^2.$ Therefore, $c^2=\cos^2\alpha+\sin^2\alpha=1.$ So, $c=1.$

Foci are at $(\pm c,0)=(\pm 1,0).$

Thus the abscissae of foci are always $\pm 1,$ which are constant.

So, B is true.

Option C: Eccentricity

Eccentricity of hyperbola is $e=\frac{c}{a}.$ Since $c=1, \ a=\cos\alpha,$ we get $e=\frac{1}{\cos\alpha}=\sec\alpha.$

This varies with $\alpha$, so it is not constant.

So, C is false.

Option D: Directrix

Directrices of the hyperbola are $x=\pm \frac{a}{e}.$ Using $e=\frac{c}{a},$ $\frac{a}{e}=\frac{a^2}{c}=a^2=\cos^2\alpha$ since $c=1.$

Hence directrices are $x=\pm \cos^2\alpha,$ which vary with $\alpha$.

So, D is false.

Conclusion

The only quantity that remains constant is the abscissae of the foci.

$\boxed{\text{Option B}}$