existence of smooth points on varieties $P_1, P_2, \ldots, P_k$ are algebraically independent irreducible homogeneous polynomials in $n$ variables over the field of complex numbers. Then, is it true that there is a point $b = (b_1, b_2, \ldots, b_n)$ such that $\forall i \in [k], P_i(b) = 0$ and the rank of the Jacobian of $P_1, P_2, \ldots, P_k$ at $b$ is equal to $k$ ? 
If this is false, a counterexample would also be very helpful.  
 A: If you furthermore suppose that the variety defined by $I=(P_1, \ldots, P_k)$ is irreducible, then the answer is yes (apart from a silly counterexample in dimension 1 which can be straightforwardly corrected). Denote by $Z(J)$ the zero locus of the ideal $J$, and $I(V)$ the ideal of polynomials which are zero over $V$. Recall that by Nustellensatz $I(Z(J) ) = \sqrt{J}$.
Infact, let $D_{(i_1\ldots i_k)}$ be the determinant of the $(i_1, \ldots, i_k)$ minor of the jacobian matrix. Take an irreducible factor $f_{(i_1 \ldots i_k)}$ of $D_{(i_1\ldots i_k)}$; the ideal generated by $f$'s are prime ideals of height 1.
Suppose by absurd that the thesis is false. Then 
$$\sqrt{I} \subseteq \bigcup_{i_1, \ldots i_k} \sqrt{(D_{(i_1, \ldots i_k)})} \subseteq \bigcup_{i_1, \ldots i_k} (f_{(i_1, \ldots i_k)}) $$
because every time $b \in Z(I)$, then $Jac_{P_i}(b)$ is not of full rank, so there is a minor which has zero determinant, so $b \in Z(D_{(i_1, \ldots i_k)})$ for some $i$'s (you get an extra $\sqrt{ }$ because of nustellensatz). 
Now if a prime is contained in a union of primes, it is containd in one of them, so $\sqrt{I} \subset (f_{(i_1 \ldots i_k)})$. If $k > 1$, you have a prime $\sqrt{I}$ (here we use the irreducibility hypothesis) of height $k$ contained in a prime of height $1$, which is an absurd. This follows from the fact that the transcendence degree of $k[x_1, \ldots, x_n]/\sqrt{I}$ is $n-k$ by the assumption on algebraic independence, so it has krull dimension $n-k$; this implies that the height of $\sqrt{I}$ is $k$, because of the formula $ht(p)+dim(A/p)$ in a domain $A$.
If $k=1$, then you should assume that $P$ is itself irreducible; this is not implied by the hypothesis of irreducibility on algebraic set because you can take powers of an irreducible and still get an irreducible variety. Otherwise $f(x)=x^2$ is a counterexample to the thesis. However, if $P$ is irreducible then $P$ and $D_i =\partial_{x_i}P$ are relatively prime, so it can'th happen that $(P) \subset (f_i)$.
Remark. The conditions $D_{(i_1 \ldots i_k)}(b) \neq 0$ are open conditions. We have seen that the open set $U \subset Z(I)$ in which this hold is non-empty. But $Z(I)$ is irreducible, so $U$ is dense: this means that indeed the jacobian is almost always of maximal rank in $Z(I)$!
