Expression and gate not gate output
inverting A'B + A'CD + C won't result in AB' + AC'D' + C'. Because \$\overline{(A+B)} = \overline{A}\ \overline{B} \ne \overline{A} + \overline{B}\$. See De Morgan's Laws for details.

Now coming to your circuit, the inputs for 3rd AND (inst5) gate are \$D\$, \$\overline{D}\$ and B hence the output of 3rd AND gate is \$D\overline{D}B = 0\$. 

So the output of OR gate (inst6) will be \$\overline{A}B + \overline{A}CD\$. reducing which you will get \$\overline{A}(B+CD)\$.

The final output will be \$\overline{(\overline{A}(B+CD))} = A+\overline{(B+CD)} = A + \overline{B}(\overline{C}+\overline{D}) = A + \overline{B}\ \overline{C} + \overline{B}\ \overline{D} \$