$f(b,c):=\sum_i^m(y_i-c-b^Tx_i)^2+\lambda b^T b$ is equivalent to $g(d,e):=\sum_i^m(y_i-e-d^T (x_i-\bar x))^2+\lambda d^T d$ under the change of variables $d=b,e=c+b^T \bar x$

ie $f(b,c)=g(b,c+b^T\bar x)$.

Therefore they have the same minimisers [same constraints on (b,c) vs (d,e)]. But this change of variables corresponds to using centred or uncentred data.

$f(b,c):=\sum_i^m(y_i-c-b^Tx_i)^2+\lambda b^T b$ is equivalent to $g(d,e):=\sum_i^m(y_i-e-d^T (x_i-\bar x))^2+\lambda d^T d$ under the change of variables $d=b,e=c+b^T \bar x$

ie $f(b,c)=g(b,c+b^T\bar x)$.

Therefore they have the same minimisers [same constraints on (b,c) vs (d,e)]. But this change of variables corresponds to using centred or uncentred data.

It should be noted that this only works when the regularisation is not on the constant term. Although regularisation is typically performed as above, some software also penalises the constant/bias term.

To prove b' =b'' you take any solution of the first problem and show that it is also minimising the second problem (b', c'') and vice versa. By adding the bt xbar to your c

$f(b,c):=\sum_i^m(y_i-c-b^Tx_i)^2+\lambda b^T b$ is equivalent to $g(d,e):=\sum_i^m(y_i-e-d^T (x_i-\bar x))^2+\lambda d^T d$ under the change of variables $d=b,e=c+b^T \bar x$

ie $f(b,c)=g(b,c+b^T\bar x)$.

Therefore they have the same minimisers [same constraints on (b,c) vs (d,e)]. But this change of variables corresponds to using centred or uncentred data.