Check Statement I

Aniline reacts with concentrated sulfuric acid first to form anilinium hydrogen sulfate, and on heating at about $453-473\,\text{K}$, rearrangement/sulfonation gives $p$-aminobenzene sulfonic acid (sulphanilic acid).

$\ce{C6H5NH2 ->[conc.\ H2SO4][453-473\ K] H2N-C6H4-SO3H}$

Now, in Lassaigne's test for sulfur, sodium fusion converts sulfur into sodium sulfide, which with sodium nitroprusside gives a violet/purple color. However, compounds containing both nitrogen and sulfur on fusion can form sodium thiocyanate, and with ferric ion this gives a blood red color due to ferric thiocyanate.

Since sulphanilic acid contains both $N$ and $S$, it gives blood red color in Lassaigne's test.

Therefore, Statement I is true.

Check Statement II

In Friedel-Crafts alkylation/acylation, aniline does not undergo the reaction normally because the lone pair on nitrogen coordinates with the Lewis acid catalyst $\ce{AlCl3}$ to form a salt/complex.

$\ce{C6H5NH2 + AlCl3 -> C6H5NH2.AlCl3}$

This interaction makes nitrogen positively polarized/effectively positively charged, reducing electron density on the benzene ring. Hence, the $-\ce{NH2}$ group loses its activating effect and behaves as a deactivating group.

Therefore, Statement II is true.

Conclusion

Both Statement I and Statement II are true.

So the correct option is:

$\boxed{\text{A}}$